aIsf continuous for all real values of x b Plot the graph of indicating the location of any discontinuities. c Find the value thatf approaches as xapproaches

Full text

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4

4

Differentiation techniques

The study of calculus allows us to describe and predict the behaviour of changing quantities such as the rate at which a disease spreads through a population.

The progress of, for example, an influenza epidemic can be studied by observing how the population of susceptible or infected persons changes over a given time interval. A graph can be drawn showing how these variables change over time and, if the equations of the graphs are known, we can use calculus to answer questions like the following:

• How do the numbers of susceptible, infected and recovered people change during the epidemic?

• When is the number of infected people the greatest? • When is the disease spreading most rapidly?

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MathsWorld Mathematical Methods Units 3 & 4

Continuity, limits and

differentiability

A good understanding of limits and continuity is essential to developing a good understanding of calculus. The concept of a limit is developed using tables of values, graphs and algebra.

Continuity

Informally, a function is said to be continuous if it can be sketched over an interval without removing pen from paper. That is, there are no breaks or holes over the interval.

Polynomial functions are examples of continuous functions. .If a function is continuous at each value of an interval,

then it is said to be continuous over the interval. .If the graph of a function is broken (or disconnected)

at x=a, then the function is said to be discontinuous at x=a.

This function with graph shown at right is continuous on any interval not containing x= 4. The graph has a hole at x= 4 because the function f is undefined there.

The graphs below are all discontinuous at x=a.

E x a m p l e

1

Consider the function f with rule .

a Is f continuous for all real values of x?

b Plot the graph of f indicating the location of any discontinuities.

c Find the value that f approaches as x approaches −1.

x

0

y

8

4

The function with rule

has a hole

(discontinuity) at x= 4. f x( ) x

2 16 –

x–4 ---=

Hole at x=a x a

y

Discontinuity

at x=a x a

c

d

y x=a

x y

Vertical asymptote

Continuous functions

A function f is said to be continuous at x =a if f(a) is defined, that is f(a) exists, and as

x approaches a from both directions, the value of f(x) approaches f(a).

f x( ) x

2

3x 2

+ +

x+1

---=

4.1

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Differentiation techniques

chapter

4

S o l u t i o n

a If x=−1 is substituted into f, a function value is not obtained on the graphics calculator screen and a table of values gives an error output as shown below.

f(−1) is undefined because you cannot divide by zero, and so the function f is discontinuous at x=−1.

b The graph of f is linear with a hole (discontinuity) at x= −1.

c As x approaches −1, the value of f approaches 1.

exercise

4.1

Note: When the domain of the function is not specified, it is assumed to be the set of all real numbers for which the function exists.

1 Sketch the graphs of the functions with rules as follows, indicating the nature and location

of any discontinuities.

a b c

GC 2.1, 2.2 CAS 2.1, 2.2

[−5, 5] by [−5, 5]

[−5, 5] by [−5, 5]

GC 2.1, 3.2 CAS 2.1, 3.2

The hole at (1, 1) can be displayed by using ZDecimal (Zoom Decimal) from the ZOOM

menu as shown below. This window is useful because Y1 is calculated for x-values increasing

by 0.1.

[−4.7, 4.7] by [−3.1, 3.1]

t i p

f x( ) x

2

2x–3

+ x–1

---= f x( ) 1

x+3

---= f x( ) 1

x+1 ( )2

---=

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2 Consider the function f with rule

a Sketch the graph of f.

b At which points of the domain of f is the function f discontinuous?

3 Explain why the following functions with domain R are continuous for all real values of x.

a f(x) = (x− 2)2 b

4 For each function with rule as follows, determine where the function is continuous.

a b

5 Consider the function f with rule

If f is continuous on R, find the value of a.

6 Consider the function with rule

If f is continuous on R, find the value of c and sketch the graph of y =f(x).

Limits

Calculating a limit involves finding the value that a function f approaches as x approaches a. If the function is continuous at x=a, then .

For example, .

If the function is not continuous at x= a, then can (in many cases) be evaluated using algebra, or it can be estimated graphically or with a table of values.

E x a m p l e

2

Consider the function with rule f(x) =x3− 3x. Find .

S o l u t i o n

As f(x) is a polynomial, it is continuous on R. Hence:

= 23− 3 × 2

= 2

E x a m p l e

3

Consider the function with rule . Investigate using:

a a numerical and graphical approach.

b an algebraic approach.

f x( ) –x x–1 ( )2 ⎩ ⎨ ⎧

= for x≤2

for x>2

f x( ) x–2 x2+4 ---=

f x( ) 1 x+4

---= f x( ) 1

x2+9 ---=

f x( )

x2–25 x–5 ---a

⎩ ⎪ ⎨ ⎪ ⎧

= if x≠ 5

if x= 5

f x( ) x+c

4–x2

⎩ ⎨ ⎧

= for x< 0

for x≥ 0

f x( ) xlim→a = f a( ) x2

xlim→2 = 8

f x( ) xa

lim

f x( )

xlim→2

x3–3x

( )

xlim→2

f x( ) = x---2+x3+x1+2

f x( )

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S o l u t i o n

a To investigate the behaviour of f as x→−1, a table of values or a graph can be used. We will investigate f as x→−1 from below (x<−1) and above (x>−1).

From below

The table of values and associated graph above both suggest that as x→−1 from below,

f(x) → 1.

We write , where the superscript label ‘–’ on x→−1– denotes x values approaching from below −1 as indicated in the tip.

From above

The table of values and graph above both suggest that as x→ 1 from above, i.e. x→−1+,

f(x) → 1 and so . Hence .

As the limit is the same from above and below the x value, we write .

b An algebraic approach can be used to calculate this limit.

=

= (factorising the numerator)

= (cancelling (x+ 1) since x≠−1)

=−1 + 2

= 1

Hence .

This confirms the result from part a.

GC 2.1, 2.2 CAS 2.1, 2.2

t i p

The following notation is commonly used to denote the left hand (from below) and right hand (from above) limits.

From left x→ −1−.

From right x→ −1+.

[−4.7, 4.7] by [−3.1, 3.1]

f x( )

xlim1– = 1

[4.7, 4.7] by [3.1, 3.1]

f x( )

xlim1+ = 1 f x( )

xlim1– = xlim1+f x( ) = 1

f x( )

xlim→–1 = 1

f x( )

xlim→–1

x2+3x+2

x+1

---xlim→–1

x+2

( )(x+1)

x+1

---xlim→–1

x+2

( )

xlim→–1

x2+3x+2

x+1

---xlim→–1 = 1

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E x a m p l e

4

Consider the function with rule . Investigate using:

a a table of values. b a graphical approach.

S o l u t i o n

a is undefined at x= 0, so we cannot use f(0) to investigate the limit. Construct a table of values to investigate from below and above x= 0.

The tables above suggest that as x→ 0, f(x) → 1.

b The graph of y=f(x) is shown at right.

The graph shows that a hole exists at (0, 1). By tracing to the left or to the right of the hole, the value of f(x) near x= 0 is close to 1. So the graph also suggests that as x→ 0, f(x) → 1.

Condition for the existence of a limit

if and only if both and .

f x( )

xlim→a = L xlimaf x( ) = L xlima+f x( ) = L

CAS 10.5

The TI-89 can be used to calculate a limit as xa from below or above x=a. The syntax

is limit(function, variable, point, [direction]). A direction of −1 indicates a limit calculation

from below (x a−) while a direction of +1 indicates a limit calculation from above (xa+).

The screenshots below illustrate these one-sided limit calculations and the actual limit calculation.

t i p

f x( ) sin x x

---= f x( )

xlim→0

sin x x

---GC 2.2 CAS 2.2

sin x x

---xlim→0

x → 0− x → 0+

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E x a m p l e

5

Consider the graph of the function f with rule

a At which points of the domain of f is the function f discontinuous?

b Find the following limits (if they exist):

i

ii

iii

S o l u t i o n

a From the graph, f(x) is discontinuous at x= 2.

b i f(x) is continuous at x= 0 so .

ii and , so does not exist.

iiif(x) is continuous at x= 3 so .

We are now in a position to define continuity in terms of limits.

x

0

y

4 3 2 1 4

3

2

1

–1

–2

f x( )

x–2 for x≤2 2x–3 for 2< <x 3

x for x≥3

⎩ ⎪ ⎨ ⎪ ⎧

=

f x( )

xlim→0

f x( )

xlim→2

f x( )

xlim→3

f x( )

xlim→0 = –2

f x( )

xlim2– = 0 xlim2+f x( ) = 1 xlim→2f x( )

f x( )

xlim→3 = 3

Continuity and limits—the connection

A function f is said to be continuous at x =a if it satisfies the following conditions:

.f(a) is defined, i.e. f(a) exists . exists

. .

A function f is said to be continuous in the interval (b, c) if it is continuous at any point

x=a in this interval.

f x( )

xlim→a

f x( )

xlim→a = f a( )

Numbersense with the spence

82

The atomic number of lead is 82 (it has 82 protons & 82 electrons). Lead is a bluish-white metal available in foil, granules, ingots, powder, rod, shot, sheet, and wire. The chemical symbol for lead, Pb is derived from Latin, plumbum and some dictionaries define a plumber as "a dealer or worker in lead" or one who installs lead pipes and fixtures. It has also been responsible for a few poisonings over the years. The band Led Zeppelin chose their name after The Who's late drummer Keith Moon told Robert Plant, Jimmy Page and the other members of the rival group that they'd "go over like a lead zeppelin".

In a mockingly ironic move, they promptly adopted the term as the name of their band.

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Endpoint continuity

Let f: [1, 2] →R where f(x) = x+ 1. The graph of y= f(x) is a straight line segment as shown. It can be sketched without lifting pen from paper, so informally the function is continuous for all points in its domain. However, the formal definition of continuity above does not apply at the endpoints, since the two-sided limit in the definition makes no sense in this case.

At the left-hand endpoint, where x= 1, the only limit that makes sense is since x itself is restricted to the

interval [1, 2] in this example. Similarly, at the right-hand endpoint where x= 2, the only limit that makes sense is .

So when a function has an endpoint, we say that in this special case, it has endpoint continuity if the appropriate one-sided limit exists and is equal to the value of the function at the endpoint. For example, the function , which starts from (0, 0), has endpoint continuity at (0, 0). In fact, it is continuous at all points in its domain.

exercise

4.1

7 Evaluate the following limits.

a b c

d e f

8 Evaluate the following limits.

a b c

9 Evaluate the following limits.

a b c

d e f

g h

10 Use a table of values or a graph to suggest a value for the following limits. [Make sure that

your calculator is set in radian mode.]

a b c

d e

x

0

y

y = x +1, 1 ≤x≤ 2

3 2 1 3 2 1

f x( ) xa

lim

f x( ) x1+

lim

f x( ) x2–

lim

y = x x, ≥0

continued

2x–3

( )

xlim→1 6–x x 2

( )

xlim→2 x

2

3–x

( )

xlim→–2

2–x

( )(6–x)

xlim→1

x+3 x–2

---xlim→1

5x–15

x+3

---xlim→0

x2

3x

---xlim→0

x+1 ( )(x+3)

x+1

---xlim→–1

x–5 ( )(x+4)

x+4

---xlim→–4

x2+x x

---xlim→0

x2–3x x

---xlim→0

x2–x x–1

---xlim→1

x2–2x–3 x–3

---xlim→3

x2–x–6 x+2

---xlim→–2

x2+3x–4 x–1

---xlim→1

x2–4x–12 x+2

---xlim→–2

2x2–7x+3

x–3

---xlim→3

1–cos x

x

---xlim→0

tan x

x

---xlim→0

sin x

x

---xlim→0

xcos 3x

sin 3x

---xlim→0

x+9 –3 x

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11 The functions listed in the table below exhibit different behaviour near x = 3. For each function, complete the table. (Use an appropriate one-sided limit in part d.)

12 Consider the function with rule .

a For what values of x is f continuous? b Find .

c Sketch the graph of f, indicating the nature and location of any discontinuities.

13 Consider the function with rule .

a For what values of x is f continuous? b Show that does not exist.

c Sketch the graph of f.

14 Consider the function with rule

a Sketch the graph of f.

b At which points of the domain of f is the function f discontinuous?

c Find the following:

i ii iii

Gradients

In MathsWorld Mathematical Methods Units 1 & 2, we found that the gradient at any point P on a curve with equation y= f(x) is given by the gradient of the tangent to the curve of P. To find the gradient of a tangent, we first find the gradient of a secant through P

and a nearby point Q as shown in the diagram on the right. As h approaches zero, Q approaches P and the secant approaches the tangent at P. Thus the gradient at P where x=a can be expressed as a limit:

provided that the limit exists.

Function f(3) Continuous at x= 3?

a

b

c

d

f x( )

xlim→3

f x( ) = x–3

f x( ) (x–3)

2

x–3

---=

f x( ) 1 x–3

---=

f x( ) = x–3

f x( ) x

2

1 – x–1 ---=

x2–1 x–1

---x→1

lim

f x( ) 1 x+2 ---=

f x( )

xlim→–2

f x( ) 3 –

2x

x+1 ⎩ ⎪ ⎨ ⎪ ⎧ =

x<–1

1≤ ≤x 1

x>1

f x( )

xlim→–2 xlim→–1f x( ) xlim→1f x( )

y

Q(a+h,f(a+h))

P(a,f(a))

Tangent Secant

Risef(a+h)−f(a)

Run

a+ha=h

a a+ h x

0

gradient at P= f a( +h)–f a( ) h

---hlim→0

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We can symbolise this using the ‘dash’ notation. Thus f′(a) is used to mean the gradient at the point where x=a on the curve with equation y=f(x). This is also called the derivative of the function with rule f(x) at the point where x=a.

E x a m p l e

6

Show that the gradient of the parabola with equation y =x2 at the point where x= 1 is 2.

S o l u t i o n

Let f(x) =x2. Then by definition:

f′(1) =

= = = 2

Local linearity

When using graphing software, ‘zooming in’ on any smooth section of the graph of a function such as a polynomial graph will cause that section to start to look like a straight line.

Consider the function with rule f(x) =x2. ‘Zooming in’ can be used to estimate the gradient of the curve at (1, 1). The two graphs below show the result of successive zooming in, and help to demonstrate that the gradient at (1, 1) is 2.

As we zoom in further, it becomes clear that the gradient at (1, 1) is 2. The more formal method used in example 6 shows that the gradient is indeed 2.

1+h

( )2

12

h

---hlim→0

1+2h+h2–1

h

---hlim→0

2+h

( )

hlim→0

x y

1.5 1.25 1 0.75 2

1.5

1

(1, 1)

0.5 Gradient at (1, 1) is 2.

Original view

Zoom in

x y

1.1 1.05 1 0.95 1.2

1.1

1

0.9

(1, 1)

≈0.2

≈0.1 Gradient at (1, 1) is 2.

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Differentiability

A function with rule f(x) is said to be differentiable at the point where x=a if it has a derivative at that point, i.e. if the graph of the function has a (non-vertical) tangent at the

point where x=a. This means that f′(a) exists, which in turn means that exists.

Some functions, like the quadratic function in example 6, or indeed any polynomial function, are differentiable at any point. Other functions are differentiable almost everywhere in their domain, except at some special points. Still other functions, really weird ones, are not differentiable anywhere (it is impossible to draw a tangent to the corresponding curve at any point)! So what type of points cause difficulties?

A function is not differentiable at x=a if the graph of the function: .is not continuous at x= a; or

.has a sharp point or cusp at x=a (a sudden change in direction, so that the graph does not have a tangent at x=a); or

.has a vertical tangent at x=a.

Roughly speaking, apart from the case of a vertical tangent, a function is differentiable wherever its graph is continuous and ‘smooth’, i.e. has no sudden changes in direction.

E x a m p l e

7

The graph of y=f(x) is shown on the right.

a Explain why the function is not differentiable at the point where x= 0.

b Check your answer to part a using the formal definition.

c What is the gradient of the function at all the points where it is differentiable?

t i p

The ‘Zoom’ factors, XFact and YFact, have values ≥1 and define the magnification or

reduction factor used to Zoom In or Zoom Out around a point. To zoom in on a graph:

.Set the XFact and YFact values by selecting SetFactors from the ZOOM menu.

.Select ZoomIn from the ZOOM menu and with the zoom cursor now displayed, move the

cursor to the point that is to be the centre of the new viewing window and press ENTER.

To return to the viewing window displayed before the previous zoom, select ZoomPrev.

GC 3.2 CAS 3.2

f a( +h)–f a( ) h

---h→0

lim

y

1

–1

x

0 1 2

–1 –2

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S o l u t i o n

a The graph is not continuous at (0, 0), so it is not differentiable there.

b If f′(0) existed, it would be given by:

f′(0) =

=

Consider the right-hand limit:

But this limit does not exist since as h→ 0+, . (Similarly the left-hand limit does not exist.)

Thus the limit does not exist, and so the function is not differentiable at (0, 0).

c At all other points, the graph consists of horizontal straight lines whose gradient is 0. So the function has gradient 0 at all points except (0, 0).

E x a m p l e

8

Consider the function f with rule and domain R. Investigate whether f is differentiable at the origin.

S o l u t i o n

The graph of y=f(x) is shown.

It appears that there is a sudden change in direction at the origin, i.e. that there is a sharp point or cusp there. This suggests that the function is not differentiable at (0, 0). We can check using the formal definition as in example 7.

f′(0) =

=

=

Consider the left-hand and right-hand limits. If h< 0, =−h, so:

= = =−1

So f is not differentiable at x= 0 as the left-hand limit is not equal to the right-hand limit. More informally, note that the graph consists of two lines: the one to the left has gradient

−1 and the one to the right has gradient 1. So no gradient exists at the point where they meet.

f(0+h)–f( )0

h

---hlim→0

f h( )–0

h

---hlim→0

f h( )–0

h

---hlim0+

1 0–

h

---hlim0+

=

1

h

---→∞

f x( ) = x

y

y = |x|

3

2

1

–1

x

0 1 2 3 –1

–2 –3

f(0+h)–f( )0

h

---hlim→0

0+h – 0

h

---hlim→0

h h

---hlim→0

If h> 0, =h, so:

= = = 1

h h h

---hlim0+

h h

---hlim0+

1

hlim0+ h

h h

---hlim0–

h

h

---hlim0–

1

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E x a m p l e

9

Consider the function f with rule and domain R. Is f differentiable at x= 0?

S o l u t i o n

The graph of y=f(x), i.e. y=x1/3, is a curve which appears to be smooth everywhere. However, at (0, 0), there may be a vertical tangent. So we check with the formal definition:

f′(0) =

=

=

However, as h→ 0, .

So the limit does not exist and hence f is not differentiable at x= 0.

E x a m p l e

1 0

Consider the function given by

a For what value of a is the function continuous for all points in (−∞, 3)?

b Sketch the graph of f for the value of a found above.

c For what values of x is f differentiable?

S o l u t i o n

a For x< 1, the graph is linear. For 1 <x< 3, the graph is parabolic. Each of these parts of the graph is continuous. For the function to be continuous everywhere in (−∞, 3), the two parts need to ‘join up’. So we look at the limit as x approaches 1.

=

=

For the limit to exist, the left-hand and right-hand limits must be equal. So a= 1.

c For x< 1, y=x and for 1 <x< 2, y= (x− 2)2. Since each of these is a special case of a polynomial, the function is differentiable for x< 1 and 1 <x< 2. Is it differentiable at the point (1, 1)?

To the left of x= 1, the gradient is 1.

Immediately to the right of x= 1, a tangent will slope down from left to right, so its gradient is negative.

Thus no gradient can exist at (1, 1) where the straight line and the parabola meet, since there is a sudden change in direction. The point (1, 1) is a cusp.

(Alternatively, we could use the formal definition to show that f′(1) does not exist.) So the function is differentiable for x∈ (−∞, 3) \ {1}.

1 5

–2 –1 –0.5

–1

–1.5

–2 2

1.5

1

0.5

2 3 4 –3

–4

–5 0 x

y

y = x1/3

1

f x( ) = x1 3⁄

f(0+h)–f( )0

h

---hlim→0

h1 3⁄ –0

h

---hlim→0

1

h2 3⁄

---hlim→0

1

h2 3⁄

---→∞

f x( ) ax x≤1

x–2

( )2

1< <x 3

⎩ ⎨ ⎧

=

(1, 1) (3, 1)

x

0

y

3 2 1 2

1 b

f x( )

xlim1– xlim1ax = a

f x( )

xlim1+ (x–2)

2

xlim1+ (1 2– )

2

1

= =

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exercise

4.1

15 Use the definition to find the gradient of the curve with

equation y =f(x) at the point given.

a f(x) = 2x2; point (1, 2).

b f(x) =−x2+ 1; point (2, −3).

c f(x) =x2 −x; point (0, 0).

16 a Sketch the graph of .

b Explain why f′(0) does not exist.

c Use the formal definition to prove that f′(0) does not exist.

17 Let f: RR where f(x) = x2/3. Determine whether f is differentiable at (0, 0).

18 In each of the following, the function has domain R and graph as given. State the values

of x for which the function is:

i continuous. ii differentiable.

Give reasons to support your answers.

a b

19 Consider the function given by .

a Sketch the graph for a= 0 and hence comment on the values of x for which the

function in this case is

i continuous. ii differentiable.

b For what value of a is f continuous on R? Is f differentiable at all points of R in this case?

20 Let f: RR where .

Determine the value of f′(0) if it exists.

continued

Continuity and differentiability

.If f is differentiable at x=a, then f is continuous at x= a.

.A function may be continuous at x=a but not differentiable at x=a. For example, the

absolute value function is continuous at x= 0 but not differentiable at x= 0.

f′( )a f a---( +hh)–f a( )

h→0

lim =

y = f x( ) = 1– x x, ∈R

x

0

y

3 2

1

−1

2 1

−1

−2

x

0

y

3 1

−1

2 1

−1

f x( ) –x–1 x+a

⎩ ⎨ ⎧

= x<0

x≥0

f x( ) x

2

x2 – ⎩ ⎨ ⎧

= x≥0

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21 Let f: RR where .

Show that there is a value for a for which f is continuous and differentiable at every

point of R.

22 a Use a graphics calculator to help sketch the graph of y= f(x) where

.

b For what values of x is f differentiable?

f x( )

x2+a

⎩ ⎨ =

x>0

f x( ) = x+1 – x + x–1,xR

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Differentiation

In section 4.1, we reviewed the use of the definition for finding the gradient of a curve with equation y= f(x) at the point where x=a. We also used this definition to determine whether a derivative existed, i.e. whether a function was differentiable at a particular point.

In this section, the focus is on rules for finding the derivative at any point x of a function with equation y =f(x), wherever the function is differentiable. The resulting derivative function is symbolised by either f′(x) or , and is also called the gradient function. The process of finding the derivative is called differentiation.

Power rule of differentiation

In Maths World Mathematical Methods Units 1 & 2, we determined that if f(x) =xn, then

f′(x) =nxn − 1 where n is a positive integer. In fact it can be shown that this result, called the power rule of differentiation, is true for all real values of n. In this subject we will restrict n

to be a rational number, i.e. nQ.

In alternative notation, if y=xn, then . More generally, if f(x) =axn, where a is a constant, then f′(x) =anxn − 1. For example, if f(x) = 3x4, then f′(x) = 4(3x3) = 12x3.

Derivative of a constant

Since f(x) =c, where c is a constant, can be written in the form f(x) =cx0 (as x0= 1), then by the power rule:

f′(x) =c(0x−1) = 0

Alternatively, the graph of y=c is a horizontal straight line with gradient 0, so at any point x on the line.

Derivative of the sum of terms

It can be shown that if h(x) =f(x) +g(x), then h′(x) =f′(x) +g′(x). In words, the derivative of a sum is equal to the sum of the derivatives.

Derivative of a polynomial

Polynomials can be differentiated one term at a time. If f(x) =anxn+ an 1x

n− 1

+ … +a2x 2

+a1x+a0, then:

f′(x) =nanxn− 1+ (n− 1)an 1x

n− 2

+ … + 3a3x 2

+ 2a2x+a1.

E x a m p l e

1

Consider the function with rule f(x) = 2x3+x2 and domain R.

a Find f′(x).

b Find the gradient of the graph of f at the point (2, 20).

f′( )a f a---( +hh)–f a( ) xlim→a

=

dy dx

---dy dx

--- = nxn–1

dy dx

--- = 0

4 . 2

(17)

S o l u t i o n

a f′(x) = 2(3x2) + 2x

= 6x2+ 2x

b Substitute x= 2 into the derivative to find f′(2).

f′(2) = 6(2)2+ 2(2)

= 24 + 4

= 28

The gradient at (2, 20) is 28.

E x a m p l e

2

Consider the function f: RR where f(x) =x3− 5x2+ 3x+ 9.

a Find f′(x).

b Find the gradient of the graph of f at the points where it meets the x-axis.

S o l u t i o n

a Differentiate one term at a time using the power rule of differentiation.

f′(x) = 3x2− 10x+ 3

b To locate where the graph meets the x-axis solve f(x) = 0 for x.

Many solution approaches exist. We could use the graphics calculator for example. The screenshots below show that x=−1, 3 are solutions to the equation f(x) = 0 and so the graph meets the x-axis at the points (−1, 0) and (3, 0).

The screenshots suggest these are the only x-intercepts. We can confirm this with algebra as follows.

Dividing by x+ 1, as f(−1) = 0, we find that:

x3− 5x2 + 3x+ 9 = (x+ 1)(x2− 6x+ 9)

= (x+ 1)(x− 3)2

So f(x) = 0 when x=−1, 3 and hence the coordinates of the x-intercepts are (−1, 0) and (3, 0).

GC 5.2 CAS 5.2

t i p

A graphics calculator can be used to calculate f′(2)

where the alternative notation, , is used. Here the

screenshot shows (the gradient) of the curve at

x= 2. Note that the derivative is calculated

numerically and so gives an approximate value.

dy dx

---dy dx

---[3, 3] by [15, 30]

GC 2.2, 2.4 CAS 2.2, 2.4

[−3, 5] by [−15, 25]

(18)

To find the gradient of the curve at (−1, 0) calculate f′(−1).

f′(−1) = 3(−1)2− 10(−1) + 3

= 3 +10 + 3

= 16

The gradient of the curve at (−1, 0) is 16.

To find the gradient of the curve at (3, 0) calculate f′(3).

f′(3) = 3(3)2− 10(3) + 3

= 27 − 30 + 3

= 0

The gradient of the curve at (3, 0) is zero and the point (3, 0) is a stationary point.

E x a m p l e

3

Consider the function with rule f(x) =−2x2− 8x− 8 and domain R. Find x such that f′(x) > 0.

S o l u t i o n

f′(x) =−4x− 8

=−4(x+ 2)

Now we need to find x such that f′(x) > 0.

−4(x+ 2) > 0

x+ 2 < 0 (dividing by a negative number reverses the inequality) x<−2

The graph has positive gradient (i.e. f′(x) > 0) for x<−2. The graph below confirms this result.

CAS 10.6

t i p

The TI-89 can be used to find derivatives and evaluate them as shown, where the function from example 2 has

been stored as f(x). The last line shows that order is

important. You must differentiate first, then substitute; since f(−1) is a constant, its derivative is zero.

f′(x) = 0 for x = −2

f′(x)<0 for x>–2

f′(x)>0 for x<–2

x

0

y

–6 –4 –2 4

2

–2

–4

–6

–8

–10

(19)

E x a m p l e

4

Consider the graph of a polynomial function f shown at right. The points labelled are points of zero gradient. Find:

a {x:f′(x) = 0}

b {x:f′(x) > 0}

c {x:f′(x) < 0}

S o l u t i o n

a {x:f′(x) = 0} = {−2, 2}

b {x:f′(x) > 0} = {x: −2 <x< 2}

c {x:f′(x) < 0} = {x:x<−2} ∪ {x:x> 2}

exercise

4.2

1 Find f′(x) for the following.

a f(x) = 5x+ 2 b f(x) = x2− 7x c f(x) =x2− 4x+ 7

d f(x) =x4 + 5x2 e f(x) = x(x− 6) f f(x) = 2x(1 −x)

g f(x) = (x− 4)2 h f(x) = (x+ 1)3 i f(x) =x(x− 2)2

2 Find f′(x) for the following.

a f(x) =ax2+bx+c b f(x) = ax3+ bx2 +cx+d c f(x) = (xb)2

3 Differentiate each of the following with respect to x.

a b

4 Consider the function with rule f(x) = x3− x. Evaluate the following.

a f′(0) b f′(1) c f′(−1) d

5 Find f′(−2) for each of the following.

a f(x) =x2 − 7x+ 6 b f(x) = 4xx3

x y

0

(2, 4)

(–2, –12)

t i p

The TI-83/84 can calculate f(a) for multiple values of a by using a table of values or at the

home screen, as shown in the second screenshot.

The TI-89 can calculate f′(a) for multiple values of a as shown in the third screenshot. Recall

that the derivative must be calculated before the substitutions. f x( ) x

2

3x

x

---,x≠0

= f x( ) 2x

3

3x2

x

---,x≠0 =

f′ 1 2 ---– ⎝ ⎠ ⎛ ⎞

GC 5.3 CAS 5.3,

10.6

(20)

6 Consider the function with rule f(x) = 2x2− 5x+ 12. Find x such that f′(x) = 3.

7 Consider the function with rule f(t) = t3− 4t. Find t such that f′(t) = 23.

8 Find the gradient of the curve with equation y= x2− 4 at the point P(3, 5).

9 Find the gradient of the curve with equation y= x2− 2x at the point P(2, 0).

10 Consider the function with rule h(x) = 4x2 + 2x. Find the coordinates of the point on the

graph of the function at which the gradient is −6.

11 Consider the function with rule g(x) = 2x3 + 3x2 + 12x+ 1. Find the coordinates of the

point(s) on the graph of the function at which the gradient is zero.

12 Find the coordinates of the point on the graph of y =x2 at which the tangent line to the

curve is parallel to the line y= 4x+ 3.

13 Find, correct to 2 decimal places, the coordinates of the point(s) where the curve with

equation y = 2x3 −x+ 4 has a gradient equal to 5.

14 Consider the function with rule f(x) = 6x+x2.

a Find f′(x).

b Find the gradient of the graph of y=f(x) at the point (−2, −8).

c Find the coordinates of the point at which the gradient is −3.

15 Consider the function with rule f(x) = x3− 3x2.

a Find f′(x).

b Find the gradient of the graph of f at the point (−1, −4).

c Find the coordinates of the point(s) at which the gradient is equal to 1.

16 Find the gradient of the curve with equation y=x2− 6x+ 5 at the point where it crosses

the y-axis.

17 Find the gradient of the curve with equation y=x2− 5x+ 6 at the points where it crosses

the x-axis.

18 Find the coordinates of the point on the curve with equation y=x2− 6x + 1 at which the

tangent is parallel to the x-axis.

19 Consider the graph whose equation is y= 4(x+ 2)2.

a Find the coordinates of the point(s) at which the graph touches the x-axis.

b Find the values of x such that the gradient is greater than 2.

20 Let f: RR where f(x) = x(x− 3)2.

a Find f′(x) and express your answer in factorised form.

b Hence find the coordinates of the points on the graph of f at which the gradient is zero.

c Find the equation of the straight line joining the points at which the graph of f has

zero gradient.

d Hence, find the point of intersection of f and the straight line that lies between the two

points of zero gradient, whose equation was found in part c. What do you notice?

21 Consider the function with domain R and rule f(x) = ax2 +bx, where a and b are real

numbers. If f(2) = 11 and f′(0) = 1, find the values of a and b.

(21)

22 Consider the function with domain R and rule f(x) = x + x. Does f′(x) ever equal zero? Explain.

23 Find the coordinates of the points on the graph of y=x2 from which the tangent line to

the graph passes through the point (3, −7).

24 Let g: RR where g(x) =−2x2− 8x− 8. Find x such that g′(x) > 0.

25 Consider the graph of a polynomial function f shown.

The points labelled are points of zero gradient. Find:

a {x:f′(x) = 0}

b {x:f(x) > 0}

c {x:f(x) < 0}

26 Consider the graph of a polynomial

function f shown.

The points labelled are points of zero gradient. Find:

a {x:f′(x) = 0}

b {x:f(x) > 0}

c {x:f(x) < 0}

27 Let f: RR where f(x) = ax2 +bx and a and b are real numbers.

a If f(1) + f(−1) = −2, find the value of a.

b If the gradient of the graph of f is zero at x= 1, find the value of b.

c Hence state the coordinates of P at which the tangent to the graph of f is parallel to the

x-axis.

d Find the equation of the tangent to the graph of f at x= 3.

e Hence or otherwise state the gradient of the graph of f at x= 3.

f Find the set of x values for which the gradient of the graph of f is less than −4.

0 x

y

(1, 3) (–1, 7)

y

x

(0, 4)

(–1, 2) (1, 2)

0

(22)

E x a m p l e

5

Find for the following.

a b

S o l u t i o n

a y=

= 1 − 2x−1

So = 0 − 2(−1)x−2

= 2x−2

=

exercise

4.2

28 Find f′(x) for the following.

a f(x) =x−1 b f(x) = 3x−2 c f(x) = 5x−4

d e f

29 Find for the following.

a b c

30 Find f′(x) for the following.

a b c

d e f

31 Find the gradient of the curve defined by at the point P(−2, 1).

32 Let . Find f′(x) and hence evaluate f′(2).

33 Find the coordinates of the point on the curve with equation at which .

34 Consider the curve with equation .

a Find the coordinates of the points where the curve crosses the x-axis.

b Find the coordinates of the point(s) where the gradient is equal to −1.

35 Let h: R \ {0} →R where . If the graph of h has zero gradient at P(1, 3),

find the values of a and b.

36 Let h: R \ {0} →R where . If the graph of g has a gradient of 8 at P(1, −2),

find the values of a and b.

dy dx

---y x–2 x

---= y 2

x

---– =

b

So =−2 x−3/2

= =

=

y = –2x–1 2⁄ dy dx --- ⎛⎜ ⎝ 1 2 ---– ⎞⎟ ⎠

x–3 2⁄

1

x–3 2⁄

---1 x x ---x x -- 2 x ---– dy dx ---2 x2

---continued

f x( ) 4 x2

---= f x( ) = x5 2⁄ f x( ) = x5 4⁄

dy dx

---y = 6 x y 1

x

---= y = 3 x

f x( ) 7x 4 1 x ---– +

= f x( ) x2–4x 2

x ---+

= f x( ) = 2xx

f x( ) x 2

x ---–

= f x( ) x 2

x ---+

⎝ ⎠

⎛ ⎞2

= f x( ) 3x x

3 – x2 ---= y 2 x ---– = f x( ) x 1

x ---+

⎝ ⎠

⎛ ⎞2

=

y = x dy

dx --- 1

8 ---=

y x–3 2

x ---+ =

h x( ) ax2 b x ---+ =

g x( ) ax

(23)

4.1Multiple derivatives

CD

SAC analysis task

The derivative of y with respect to x is written as . If we take the derivative of with respect to x, we have the second derivative of y with respect to x and this is written as

. The third derivative of y with respect to x is written as , and so on.

a If y= x10, find

i ii iii

b If y= x10, show that

(The coefficient of x10 −n is the product of a sequence of descending numbers starting at 10 and with the last one being (11 −n).)

c Hence find .

d Find if

i y =x5 ii y=x7 iiiy= x12

e Find if y= xn and

i n= 4 ii n= 5 iii n= 6 iv n= 100

dy dx

--- dy

dx

---d2y dx2

--- d

3 y dx3

---dy dx

--- d

2 y dx2

--- d

3 y dx3

---dny dxn

--- = 10×9×… 11( –n)x10–n

d10y dx10

---d3y dx3

---dny dxn

---analysis task 1—

multiple derivatives

SAC

Numbersense with the spence

83

83% of people hit by lightning are men. The most anyone has been hit by

lightning is 7 times. The lucky recipient avoided an 8th by committing suicide before it could happen.

(24)

4.2Derivatives of ‘nice’ cubic polynomials

CD

SAC analysis task

The findings from this task can assist your teacher in their choice of future calculus problems. In doing worked examples for students studying calculus, mathematics teachers frequently want to use a cubic polynomial defined by a rule f(x) that has the following properties:

.The equation f(x) = 0 has solutions that are rational numbers.

.The equation f′(x) = 0 also has solutions that are rational numbers.

Rational numbers are numbers that can be written in the form where a and b are integers.

are examples of rational numbers.

We will call cubic polynomials that possess the above two properties ‘nice’ cubic polynomials.

Consider the function defined by the rule f(x) =x(xpr)(xqs) where p, q, r and s are rational numbers.

a For the case where p = 1, q= 3, r= 5 and s= 7:

i solve f(x) = 0 for x. ii find f′(x). iiisolve f′(x) = 0 for x. ivstate whether this cubic polynomial is ‘nice’.

b For the case where p =−2, q= 1, r= 4 and s = 7:

i solve f(x) = 0 for x. ii find f′(x). iiisolve f′(x) = 0 for x. ivstate whether this cubic polynomial is ‘nice’.

c What conditions must be placed on p, q, r and s for a cubic polynomial to be ‘nice’? d Repeat part a with a suitably chosen set of values for p, q, r and s that

i satisfies the conditions for a ‘nice’ cubic polynomial.

ii does not satisfy the conditions for a ‘nice’ cubic polynomial.

e For a ‘nice’ cubic polynomial, describe how the solutions to the equations f(x) = 0 and

f′(x) = 0 are formed from the set of rational numbers p, q, r and s.

f Construct a ‘nice’ cubic polynomial defined by the rule f(x) =ax3+bx2+cx+d where

a, b, c and d are real numbers and f(0) = 0, f(5) = 0, f′(3) = 0 and f′(−1) = 8. g Consider the ‘nice’ cubic polynomial with rule f(x) =x(x− 5)(x− 21)

i Show that f(x) = x(x+ 5)(x+ 21) is also a ‘nice’ cubic polynomial. ii Show that f(x+ 1) is a ‘nice’ cubic polynomial.

iii Show that f(−x) is a ‘nice’ cubic polynomial. iv Show that f(3x) is a ‘nice’ cubic polynomial. v Generalise your results from above.

h For the case where p =tk, q=t, r=t+k and s= t+ 2k and t and k are rational numbers, show that f(x) is a ‘nice’ cubic polynomial.

a b

---1 4

--- 3 2 5

---– –4 and 17 3

---, ---, , ,

analysis task 2—

(25)

Differentiation rules

The chain rule

Let y= (x2− 1)3. To find , we can expand (x2− 1)3 and then differentiate as shown below.

= (perfect cubic expansion)

=

So .

Factorising leads to an interesting connection between y and in this instance.

6x5− 12x3 + 6x= 6x(x4 − 2x2 + 1)

= 6x(x2 − 1)2

= 3(2x)(x2− 1)3 − 1

It appears that the following changes have been made to the original expression on the right-hand side.

Let’s look at this in another way. Consider the function whose rule is y = (x2− 1)3. It is an example of a composite function, since it is a function ‘composed’ of two functions.

Another way of representing a composite function is as a ‘chain’ of two functions.

If we let the inside function be denoted by u, then where u=x2− 1. To find , first find and .

and .

dy dx

---x21

( )3

x2 ( )3

3 1( )( )x2 2

+3 1( )2( )x2 ( )1 3 x6–3x4+3x2–1

dy dx

--- = 6x5–12x3+6x dy

dx

--- dy

dx

---dy dx

--- = ( )3 ( )2x (x2–1)3–1

The power of the original expression comes out the front

One is subtracted from the power of the original function

The derivative of the function appears inside the bracket

outside function

inside function

y = (x21) 3

function 1

‘square and subtract one’

function 2

cube input

x x21 (x2 1)3

y = (x21)3 = u3 dy

dx

--- dy

du

--- du

dx

---dy du

--- = 3u2 du dx

--- = 2x

4.3

4 . 3

(26)

Now multiply and simplify:

=

= 3u2× 2x = (3)(2x)u2 = (3)(2x)(x2− 1)2

= 6x(x2− 1)2

This chain rule method is an efficient way of finding the derivative of a composite function. Its rule is stated more formally below.

For example to find , given , proceed as follows.

Let u=x2− 1 and so y= u10.

= 2x and

= = =

Imagine trying to differentiate by first expanding the expression on the right-hand side!

E x a m p l e

1

Find the gradient of the curve with equation at the point (3, 3).

S o l u t i o n

Let u= 2x+ 3 and so as .

= 2 and

=

=

=

At x= 3, and so

the gradient at (3, 3) is .

dy dx

--- dy

du

--- du

dx

---×

Chain rule

If y=f(u) and u=g(x), then .

Alternatively, using function notation, the chain rule states that: if h(x) =f[g(x)], then h′(x) =f′[g(x)]g′(x)

dy dx

--- dy

du

--- du

dx

---=

dy dx

--- y = (x21)10

du dx

--- dy

du

--- = 10u9 = 10(x21)9 dy

dx

--- dy

du

--- du

dx

---10(x21)9×( )2x

20x x( 21)9

y = 2x+3

t i p

The TI-89 can be used to illustrate the chain rule differentiation for a ‘function of a function’ situation as shown. CAS 10.6

y = u1 2⁄ u = u1 2⁄ du

dx

--- dy

du

--- 1

2

---u–1 2⁄

=

dy dx

--- dy

du

--- du

dx

---2 1 2

---u–1 2⁄

×

1 2x+3

---dy dx

--- 1

2 3( )+3

--- 1

3

---= =

1 3

(27)

---E x a m p l e

2

The table gives information about functions

f and g and their derivatives f′ and g′.

Given that h(x) =f[g(x)], find the value of h′(1).

S o l u t i o n

Use the chain rule with function notation h′(x) =f′[g(x)]g′(x).

h′(1) =f′[g(1)]g′(1)

=f′(0)g′(1) (as g(1) = 0)

= (−3)(−2) (as f′(0) =−3 and g′(1) =−2)

= 6

exercise

4.3

1 Use the chain rule to differentiate each of the following with respect to x.

a (3x2+ 1)3 b (x− 2)4 c (x3− 1)6 d

e (6 − x)4 f g h

2 Differentiate each of the following with respect to x.

a (x2− 5x + 1)3 b (x3+ x2− 7x)5 c d

3 Find the exact gradient of the curve with equation at the point

.

4 Find the gradient of the curve with equation y= (x2− 4)3 at the point where x= 2.

5 Find the gradient of the curve with equation y= (x2− 5x+ 6)2 at the points where it meets

the x-axis.

6 Let f: RR where f(x) = 4(x+ 2)4.

a State the range of f.

b Find the x values for which the gradient of the graph of f is greater than . Express

your answer correct to two decimal places.

7 Consider two curves whose equations are y= and y= respectively.

a Find the coordinates of the point of intersection for these two curves.

b Show that these curves have the same gradient at the point of intersection.

8 Consider the function g: DR where g(x) = .

a Find the set D, given that it represents the maximal domain.

b Find g′(x).

c For what real values of x is g′(x) defined?

d Find the coordinates of the point of zero gradient on the graph of g.

e Sketch the graph of g, and identify its key features.

x 0 1 2

f(x) 3 −2 1

f′(x) −3 1 4

g(x) 2 0 1

g′(x) 2 −2 3

1 2

x ---–

⎝ ⎠

⎛ ⎞4

1

3x2+5

( )4

--- x–1 x2+1

4 x2+3x+5

--- 3 x–1

y = 9–x2,–3≤ ≤x 3

2, 5

( )

1 2

---2x+1 x(4–x)

x2

– +x+12

(28)

9 The table gives information about functions

f and g and their derivatives f′ and g′.

Given that h(x) = f[g(x)], find the value

of h′(1).

10 Consider the function with rule f(x) = .

a State the implied domain and give the range of the function f.

b Find the set of x values such that the gradient of the curve with equation y= f(x) is

greater than . Express your answer correct to two decimal places.

11 Let f: RR be a differentiable function. For all real values of x, show that the derivative

of f(e2x) with respect to x is 2e2x(f′(e2x)).

12 Suppose that h(x) =f(g(x)) and g(4) = 8, g′(4) = 6, f′(4) = 4 and f′(8) = 3. Find the value

of h′(4).

The product rule

The derivative of a product is not simply the product of the derivatives. For example, suppose

f(x) = 2x5 and g(x) =x2 − 1 and h(x) =f(x)g(x).

h(x) = 2x5(x2− 1)

= 2x7− 2x5

The derivative of the product function is:

h′(x) = 14x6 − 10x4 = 2x4(7x2− 5)

The derivatives of f(x) and g(x) are:

f′(x) = 10x4 and g′(x) = 2x

The product of the derivatives is:

f′(x)g′(x) = (10x4) (2x)

= 20x5

Note that h′(x) ≠f′(x)g′(x).

It can be shown that if h(x) =f(x)g(x), then the following rule, commonly known as the product rule, applies.

x 0 1 2

f(x) 1 −1 3

f′(x) −1 0 1

g(x) −2 2 1

g′(x) −3 2 −1

x2+4

1 2

---Product rule

If h(x) =f(x)g(x) then h′(x) =f(x)g′(x) +g(x)f′(x).

In alternative notation, if u=f(x) and v=g(x), the product rule can be expressed as: , or (uv)′=uv′ + vu

d dx

---( )uv u dv dx

--- v du dx

(29)

E x a m p l e

3

a Find for each of the following.

i y= (x− 2)(x2− 1) ii

b Find the value of x at which the curve with equation in part ii has zero gradient.

S o l u t i o n

a i Since this is a product of two functions, use the product rule to find . Let u=x− 2 and v=x2− 1.

So and .

=

= (x− 2)(2x) + (x2− 1)(1)

= 2x2− 4x+x2− 1

= 3x2− 4x− 1

ii Let u=x and .

So , and to find , first use the chain rule.

Thus .

Now apply the product rule.

=

=

=

b Zero gradient means .

Express as a single algebraic fraction so that

solving for x becomes an easier task.

=

=

=

So where 8 − 3x= 0 (if , then a= 0). Thus .

dy dx

---y = x 4–x

dy dx

---du dx

--- = 1 dv

dx

--- = 2x dy

dx

--- u dv dx

--- v du dx

---+

v = 4–x du

dx

--- = 1 dv

dx ---dv dx --- 1 2 ---⎝ ⎠

⎛ ⎞( )1 (4x)–1 2⁄ –1 2 4–x

---= =

dy dx

--- u dv dx

--- v du dx

---+

x

( ) –1

2 4–x

---⎝ ⎠

⎛ ⎞+( 4x)( )1

4–x x

2 4–x

---–

CAS 10.4

t i p

With a TI-89, we can use the comDenom command to transform the derivative into a single algebraic fraction as shown.

dy dx

--- = 0

dy dx

---dy dx

--- = 0

4–x x

2 4–x

---– (2 4–x)( 4–x)–x

2 4–x

---2 4( –x)–x

2 4–x

---8 3– x

2 4–x

---dy dx

--- = 0 a

b

--- = 0

x 8

3

---=

(30)

E x a m p l e

4

Consider the function whose rule is f(x) =x2g(x). If g(2) = 5 and g′(2) =−3, find f′(2).

S o l u t i o n

Using the product rule gives:

f′(x) =x2g′(x) + 2x g(x) Substituting x= 2 gives:

f′(2) = 4g′(2) + 4g(2)

= (4)(−3) + (4)(5)

=−12 + 20

= 8

exercise

4.3

13 Differentiate each of the following with respect to x.

a (x+ 2)(x− 3) b (x+ 4)(x− 1)2 c (x2 −x− 2)(x2+ 3)

d 2x(3 −x)3 e f

14 Consider the function f with rule .

a Find f′(x) and express it in the form .

b State the value of k.

c For what values of x will the gradient of f be positive?

15 Find the gradient of the curve with equation y= x(x− 2)3 at the point (1, −1).

16 Consider the function f with rule f(x) =x(xk)2 where k is a real number.

a Find f′(x) and express your answer in factorised form.

b Find the coordinates of the points at which the graph of f has zero gradient.

c Show that the equation of the straight line joining the points where the graph of f has

zero gradient is .

d Hence show that the point of intersection of f and the equation of the straight line

found in part c occurs at the midpoint between the two points of zero gradient.

17 Consider the function with rule f(x) = xg(x). If g(−1) = 3 and g′(−1) = −2, find f′(−1).

18 Consider the function with rule f(x) = x3g(x). If g(1) = 1 and g′(1) =−2, find f′(1).

19 Consider the curve with equation y= (x− 1)(x+ 2)2.

a Find .

b Find the exact coordinates of the points on the curve where .

c Find, correct to 2 decimal places, the coordinates of the points on the curve where

.

t i p

The TI-89 can be used to start example 4 as shown.

CAS 10.1, 10.6

continued

x x2+1 x

x+1 ( )2

---f x( ) = x 3–x

k(2–x)

2 3–x

---y 2k

2

– 9

---(xk) =

dy dx

---dy dx --- = 0

Figure

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References

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