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19
Differentiation of
Polynomials
Objectives
To understand the concept oflimit.
To understand thedefinition of differentiation.
To understand and use the notation for thederivativeof a polynomial function.
To be able to find thegradientof a curve of a polynomial function by calculating
its derivative.
Toapplythe rules for differentiating polynomials to solving problems.
In the previous chapter the rate of change of one quantity with respect to a second quantity has been considered. In this chapter a technique will be developed for calculating the rate of change for polynomial functions. To illustrate this an introductory example will be considered.
On planetX, an object falls a distance ofymetres intseconds, wherey=0.8t2. (Note:On Earth the commonly used model isy=4.9t2.)
Can a general expression for the speed of such an object aftertseconds be found? In the previous chapter it was found that the gradient of the curve at a given pointPcan be approximated by finding the gradient of a chordPQ, whereQis a point on the curve as close as possible toP.
y
y = 0.8t2
t (5 + h)
Q
P
5 0
The gradient of chordPQapproximates the speed of the object atP. The
‘shorter’ this chord is made the better the approximation.
LetPbe the point on the curve wheret=5. LetQbe the point on the curve corresponding tohseconds aftert=5, i.e.Qis the point on the curve wheret=5+h.
The gradient of chordPQ= 0.8
(5+h)2−52 (5+h)−5
= 0.8
(5+h)2−52 h
=0.8(10+h)
The table gives the gradient for different values ofh. Use your calculator to check these.
If values ofhof smaller and smaller magnitude are taken, it is found that the gradient of chordPQgets closer and closer to 8. At the point wheret=5 the gradient is 8. Thus the speed of the object at the momentt=5 is 8 metres per second.
The speed of the object at the momentt=5 is the limiting value of the gradients ofPQasQapproachesP.
h Gradient ofPQ
0.7 8.56
0.6 8.48
0.5 8.40
0.4 8.32
0.3 8.24
0.2 8.16
0.1 8.08
A formula for the speed of the object at any timetis required. LetPbe the point with coordinates (t, 0.8t2) on the curve andQbe the point with coordinates (t+h, 0.8(t+h)2).
The gradient of chordPQ= 0.8
(t+h)2−t2 (t+h)−t
=0.8(2t+h)
From this an expression for the speed can be found. Consider the limit ashapproaches 0, that is, the value of 0.8(2t+h) ashbecomes arbitrarily small.
The speed at timetis 1.6tmetres per second. (The gradient of the curve at the point corresponding to timetis 1.6t.) This technique can be used to investigate the gradient of similar functions.
19.1
The gradient of a curve at a point, and the
gradient function
Consider the functionf:R→R,f(x)=x2. The gradient of the chordPQin the adjacent graph
= (a+h)2−a2
a+h−a
= a2+2ah+h2−a2
a+h−a
=2a+h
and the gradient atPcan be seen to be 2a. The limit of (2a+h) ashapproaches 0 is 2a. f (x) = x2
y
(a + h, (a + h)2) Q
P(a, a2) x
0
The gradient of the functiony=x2at any pointxis equal to 2x.
It is said that 2xisthe derivative ofx2with respect toxor, more briefly,the derivative of x2 is 2x.
This itself is clearly the rule for a function ofxand we refer to this function as thegradient (orderived) function.
The straight line that passes throughPand has gradient 2ais called thetangentto the curve atP.
From the discussion at the beginning of the chapter it can be seen that the derivative of 0.8t2 is 1.6t.
Example 1
By first considering the gradient of the chordQP, find the gradient ofy=x2−2xat the point
Qwith coordinates (3, 3).
Solution
Consider chordPQ: Gradient ofPQ
= (3+h)2−2(3+h)−3
3+h−3
= 9+6h+h2−6−2h−3
3+h−3
= 4h+h2
h
=4+h
P
2
0 x
y y = x2 – 2x
Q(3, 3) (3 + h, (3 + h)2 – 2(3 + h))
Considerhas it approaches zero. The gradient at the point (3, 3) is 4.
Example 2
Find the gradient of chordPQand hence the derivative ofx2+x.
Q(x, x2 + x)
y
x
y = x2 + x
0 –1
Solution
The gradient of chordPQ= (x+h)
2+(x+h)−(x2+x)
x+h−x
= x2+2x h+h2+x+h−x2−x
h
= 2x h+h2+h
h
=2x+h+1
From this it is seen that the derivative ofx2+xis 2x+1.
The notation for limit of 2x+h+1 ashapproaches 0 is lim
h→02x+h+1.
The derivative of a function with rule f(x) may be found by first finding an expression for the gradient of the chord fromQ(x, f(x)) toP(x+h, f(x+h)) and then finding the limit of this expression ashapproaches 0.
Example 3
By first considering the gradient of the chord fromQ(x, f(x)) toP(x+h, f(x+h)) for the
curve f(x)=x3, find the derivative ofx3.
Solution
f(x)=x3
f(x+h)=(x+h)3
The gradient of chordPQ= f(x+h)− f(x) (x+h)−x
= (x+h)3−x3
(x+h)−x
The derivative of f(x)= lim h→0
(x+h)3−x3 (x+h)−x
= lim h→0
x3+3x2h+3h2x+h3−x3
h
= lim h→0
3x2h+3h2x+h3 h
= lim h→03x
2+3hx+h2
=3x2
Example 4
Find: a lim
h→022x
2+20x h+h b lim
h→0
3x2h+2h2 h c lim
h→03x d hlim→04
Solution
a lim h→022x
2+20x h+h=22x2
b lim h→0
3x2h+2h2
h =hlim→03x
2+2h
=3x2 c lim
h→03x =3x
d lim h→04=4
Using a CAS calculator
For Example 4b, select3:limit(from theCalc, F3 , menu and complete as shown.
Exercise
19A
1 A space vehicle moves so that the distance travelled over its first minute of motion is given byy=4t4, whereyis the distance travelled in metres andtthe time in seconds. By finding the gradient of the chord between the points wheret=4 andt=5, estimate the speed of the space vehicle whent=5.
2 A population of insects grows so that the population,P, at timet(days) is given by
P=1000+t2+t, wheret>0. By finding the gradient of the chord between the points
wheret=3 andt=3+h, find an estimate for the rate of growth of the insect population at timet=3.
3 Find:
Example4
a lim h→0
2x2h3+x h2+h
h b hlim→0
3x2h−2x h2+h
h c hlim→020−10h
d lim h→0
30hx2+2h2+h
4 Find:
Example3
a lim h→0
(x+h)2+2(x+h)−(x2+2x)
h i.e. the derivative ofy=x 2+2x
b lim h→0
(5+h)2+3(5+h)−40
h i.e. the gradient ofy=x
2 +3x, wherex=5
c lim h→0
(x+h)3+2(x+h)2−(x3+2x2)
h i.e. the derivative ofy=x 3+2x2
5 For a curve with equationy=3x2 −x:
Example1
a find the gradient of chordPQ, wherePis the point (1, 2) andQis the point
((1+h), 3(1+h)2−(1+h))
b find the gradient ofPQwhenh=0.1 c find the gradient of the curve atP.
6 For a curve with equationy= 2 x:
a find the gradient of chord ABwhereAis the point (2, 1) andBis the point (2+h), 2
2+h
b find the gradient ofABwhenh=0.1 c find the gradient of the curve atA.
7 For a curve with equationy=x2+2x−3:
a find the gradient of chordPQ, wherePis the point (2, 5) andQis the point ((2+h), (2+h)2+2(2+h)−3)
b find the gradient ofPQwhenh=0.1 c find the gradient of the curve atP.
19.2
The derived function
In Section 19.1 we saw how the gradient function of a function with rule f(x) could be derived by considering the gradient of a chordPQon the curve ofy= f(x).
Consider the graphy= f(x) of the functionf:R→R.
The gradient of the chordPQ= f(x+h)− f(x)
x+h−x
= f(x+h)− f(x)
h
Therefore the gradient of the graph atPis given by
lim h→0
f(x+h)− f(x)
h
Q
P (x, f(x))
(x + h, f(x + h)) y = f(x) y
The reader is referred to Section 19.4 for a further discussion of limits.
The gradient or derived function is denoted by f, where f:R→Rand f(x)= lim
h→0
f(x+h)− f(x)
h
In this chapter only polynomial functions are considered. For a polynomial function the derived function always exists and is defined for every number in the domain off.
Determining the derivative of an expression or the derived function by evaluating the limit is calleddifferentiation by first principles.
Example 5
For f(x)=x2+2xfind f(x) by first principles.
Solution
f(x)= lim
h→0
f(x+h)− f(x)
h
= lim h→0
(x+h)2+2(x+h)−(x2+2x) h
= lim h→0
x2+2x h+h2+2x+2h−x2−2x
h
= lim h→0
2x h+h2+2h h
= lim
h→02x+h+2
=2x+2
∴ f(x)=2x+2
Example 6
For f(x)=2−x3find f(x) by first principles.
Solution
f(x)= lim
h→0
f(x+h)− f(x)
h
= lim h→0
2−(x+h)3−(2−x3) h
= lim h→0
2−(x3+3x2h+3x h2+h3)−(2−x3) h
= lim h→0
−3x2h−3x h2−h3 h
= lim h→0−3x
2−3x h−h2
Using a CAS calculator
Define f(x)=2−x3.Then, from theCalcmenu selectA:nDeriv(f(x)), where f(x) is defined as 2−x3, and take the limit ash→0.
The following results have been obtained: For f(x)=x, f(x)=1.
For f(x)=x2, f (x)=2x. For f(x)=x3, f (x)=3x2. For f(x)=x4, f (x)=4x3. For f(x)=1, f(x)=0.
This suggests the following general result:
For f(x)=xn, f(x)=nxn−1,n=1, 2, 3, . . . and for f(x)=1, f(x)=0
From the previous section it can be seen that fork, a constant:
If f(x)=kxn,the derivative function f has rule f(x)=knxn−1. It is worth making a special note of the results:
Ifg(x)=k f(x),wherek is a constant, theng(x)=k f(x). That is,the derivative of a number multiple is the multiple of the derivative. For example, forg(x)=5x2, the derived functiong(x)=5(2x)=10x.
Another important rule for differentiation is:
If f(x)=g(x)+h(x), then f(x)=g(x)+h(x). That is,the derivative of the sum is the sum of the derivatives.
Example 7
Find the derivative ofx5−2x3 +2, i.e. differentiatex5−2x3+2 with respect tox.
Solution
f(x)=x5−2x3+2
then f (x)=5x4−2(3x2)+2(0)
=5x4−6x2
Using a CAS calculator
Select1:d(from theCalcmenu ( F3 ) and complete as shown.Example 8
Find the derivative of f(x)=3x3 −6x2+1 and f(1).
Solution
f(x)=3x3−6x2+1
then f (x)=3(3x2)−6(2x)+1(0)
=9x2−12x
f (1)=9−12
= −3
Using a CAS calculator
Define f(x)=3x3−6x2+1.Example 9
Find the gradient of the curve determined by the rule f(x)=3x3−6x2+1 at the point
(1,−2).
Solution
Now f(x)=9x2−12xand f(1)=9−12= −3. The gradient of the curve is−3 at the point (1,−2).
An alternative notation for the derivative is the following:
Ify=x3,then the derivative can be denoted by d y
d x,so that d y
d x =3x
2.
In general, ifyis a function ofx, the derivative ofywith respect toxis denoted by d y
d x and
with the use of different symbolsz, wherezis a function oft. The derivative ofzwith respect
totis d z dt.
In this notationdis not a factor and cannot be cancelled. This came about because in the eighteenth century the standard diagram for finding the limiting gradient was labelled as in the figure. (is the lower case Greek letter for ‘d’, and is pronounceddelta.)
xmeans a difference inx.
ymeans a difference iny.
Q
P y
δy δx
x
0
Example 10
a If y=t2,find d y
dt. b Ifx =t
3+t,find d x
dt. c Ifz= 1 3x
3+x2,find d z d x.
Solution
a y=t2 then d y
dt =2t
b x =t3+t thend x
dt =3t
2+1
c z = 1 3x
3+x2
then d z
d x =x
2+ 2x
Example 11
a For y=(x+3)2,find d y
d x. b For z=(2t−1)
2(t+2),find d z dt. c For y= x
2+3x x ,find
d y
d x. d Differentiatey=2x
Solution
a It is first necessary to write y=(x+3)2in expanded form.
∴ y=x2+6x+9
and d y
d x =2x+6
b Expanding:
z=(4t2−4t+1)(t +2)
=4t3−4t2+t+8t2−8t+2
=4t3+4t2−7t+2
and d z
dt =12t
2+8t−7
c First divide byx:
y= x+3
∴ d y
d x =1
d y =2x3−1
∴ d y
d x =6x
2
Operator notation
‘Find the derivative of 2x2−4xwith respect tox’ can also be written as d
d x(2x
2−4x),and, in
general, d
d x(f(x))= f
(x).
Example 12
Find:
a d
d x(5x−4x 3
) b d
d z(5z 2−
4z) c d
d z(6z 3−
4z2)
Solution
a d
d x(5x−4x
3)
=5−12x2
b d
d z(5z
2−4z)
=10z−4
c d
d z(6z
3−4z2)
=18z2−8z
Example 13
Find the coordinates of the points on curves determined by each of the following equations at which the gradient has the given value:
a y=x3; gradient=8 b y=x2−4x+2; gradient=0
Solution
a y =x3implies d y
d x =3x
2
∴3x2 =8
∴ x = ±
8 3 =
±2√6 3
∴coordinates are
2√6 3 ,
16√6 9
and
−2√6 3 ,
−16√6 9
b y=x2−4x+2 implies d y
d x =2x−4
∴2x−4=0
∴ x =2
∴coordinates are (2,−2)
c y =4−x3implies d y
d x = −3x
2
∴−3x2 = −6
∴ x2 =2
∴ x = ±√2.
∴ coordinates are
2 1 2,4−2
3 2
and
−2 1 2,4+2
3 2
Using a CAS calculator
Define f(x)=4−x3. Take the derivative and store asd f(x). Now solve the equation d f(x)= −6. Substitute in f(x) to find they-coordinates.
Exercise
19B
1 For each of the following, find f(x) by finding lim h→0
f(x+h)− f(x)
h :
Examples5, 6
a f(x)=3x2 b f(x)=4x c f(x)=3 d f(x)=3x2 +4x+3 e f(x)=2x3−4 f f(x)=4x2−5x g f(x)=3−2x+x2
2 Find the derivative of each of the following with respect tox.
Example7
a x2+4x b 2x+1 c x3−x
d 1 2x
2−
3 For each of the following find f(x):
a f(x)=x12 b f(x)=3x7 c f(x)=5x d f(x)=5x+3 e f(x)=3 f f(x)=5x2−3x g f(x)=10x5+3x4 h f(x)=2x4− 1
3x 3−1
4x 2+
2
4 For each of the following, find d y d x:
Example10
a y= −x b y=10 c y=4x3−3x+2 d y= 1
3(x
3−3x+6) e y=(x+1)(x+2) f y=2x(3x2−4)
g y= 10x 5+3x4 2x2 ,x =0
5 a For the curve with equationy=x3 +1 find the gradient at points:
Example9
i (1, 2) ii (a,a3+1) b Find the derivative ofx3 +1 with respect tox.
6 a Given thaty=x3−3x2+3x, find d y
d x.Hence show that d y
d x ≥0 for allx, and interpret this in terms of the graph ofy=x3−3x2+3x.
b Given thaty= x 2+2x
x ,forx=0, find d y d x. c Differentiatey=(3x+1)2with respect tox.
7 At the points on the following curves corresponding to the given values ofx, find the y-coordinate and the gradient.
a y=x2−2x+1,x=2 b y=x2+x+1,x=0 c y=x2−2x,x= −1 d y=(x+2)(x−4),x=3 e y=3x2−2x3,x= −2 f y=(4x−5)2,x = 1
2
8 a For each of the following, find f (x) and f (1), ify= f(x) then find the
{(x,y):f(x)=1}i.e. the coordinates of the points where the gradient is 1. i y=2x2−x ii y=1+ 1
2x+ 1 3x
2
iii y=x3+x iv y=x4 −31x
b What is the interpretation of{(x,y):f(x)=1}in terms of the graphs? 9 Find:
Example12
a d dt(3t
2−4t) b d
d x(4−x
2+x3) c d
d z(5−2z
2−z4)
d d d y(3y
2−y3)
e d
d x(2x
3−
4x2) f d
dt(9.8t 2−
10 Find the coordinates of the points on the curves given by the following equations at which
Example13
the gradient has the given values:
a y=x2; gradient=8 b y=x3; gradient=12
c y=x(2−x); gradient=2 d y=x2−3x+1; gradient=0
e y=x3−6x2 +4; gradient= −12 f y=x2−x3; gradient= −1
19.3
Graphs of the derived or gradient function
Consider the gradient for different intervals of the graph ofy=g(x) shown opposite.
At a point (a,g(a)) of the graphy=g(x) the gradient isg(a).
y
y = g(x)
b
a
x R(b, g (b))
0
S(a, g(a)) Some of the features of the graph are:
x<bthe gradient is positive, i.e.g(x)>0
x=bthe gradient is zero, i.e.g(b)=0
b<x<athe gradient is negative, i.e.g(x)<0 x=athe gradient is zero, i.e.g(a)=0
x>athe gradient is positive, i.e.g(x)>0
Example 14
For the graph off:R→R, find:
a {x: f(x)>0} b {x: f(x)<0} c {x: f(x)=0}
Solution
a {x: f(x)>0}={x:−1<x<5}=(−1, 5) b {x: f(x)<0}={x:x<−1}∪{x:x>5}
=(−∞,−1)∪(5,∞) c {x: f(x)=0}={−1, 5}
0
(5, 6)
(–1, –7) y
x
y = f(x)
Example 15
Sketch the graph ofy= f(x) for each of the following. (It is impossible to determine all
features.) a
0 2
(3, –1) 4
x y
y = f(x)
b y
y = f(x)
x 0
–1 1
c
(–1.5, 4)
–3 –1 0 4
(1, – 4) y = f(x)
Solution
a f(x)>0 forx >3 f(x)<0 forx <3 f(x)=0 forx =3
y
y = f '(x)
x 0 3
b f (x)=1 for allx
c
y
x y = f ′(x) 1
0
y = f ′(x) y
x 1 −1.5 0 f(x)>0 forx>1
f(x)<0 for −1.5<x <1 f(x)>0 forx<−1.5 f(−1.5)=0 andf(1)=0
An angle associated with the gradient
of a curve at a point
The gradient of a curve at a point is the gradient of the tangent at that point. A straight line, the tangent, is associated with each point on the curve.
Ifis the angle a straight line makes with the positive direction of thex-axis, then the gradient,m, of the straight line is equal to tan, i.e., tan=m.
If=45◦then tan=1 and the gradient is 1.
If=20◦then the gradient of the straight line is tan 20◦. If=135◦then tan= −1 and the gradient is−1.
Example 16
Find the coordinates of the points on the curve with equationy=x2−7x+8 at which the
tangent:
a makes an angle of 45◦with the positive direction of thex-axis
b is parallel to the liney= −2x+6.
Solution
a d y
d x =2x−7
2x−7=1 (tan 45◦=1) and 2x=8
∴ x=4
y=42−7×4+8= −4
coordinates are (4,−4)
b Note: y= −2x+6 has gradient−2
∴ 2x−7= −2
and 2x=5
∴ x=5
2
coordinates are
5 2,−
13 4
Example 17
The planned path for a flying saucer leaving a planet is defined by the equation
y= 1 4x
4+2 3x
3 for x >0.
The units are kilometres. (Thex-axis is horizontal and they-axis vertical.)
a Find the direction of motion when thex-value is:
i 2 ii 3
b Find a point on the flying saucer’s path where the path is inclined at 45◦to the positive
x-axis. (i.e. where the gradient of the path is 1).
c Are there any other points on the path which satisfy the situation described in part b?
Solution
a d y
d x =x
3+2x2
i Whenx =2,d y
d x =8+8=16
tan−116=86.42◦ (to thex-axis)
ii Whenx=3,d y
d x =27+18=45
tan−145=88.73◦(to thex-axis) b, c When the flying saucer is flying at 45◦to the direction of thex-axis, the gradient
of the curve of its path is given by tan 45◦.
Thus to find the point at which this happens we consider equation d y
d x =tan 45
◦.
∴ x3+2x2=1
∴ x3+2x2−1=0
∴ (x+1)(x2+x−1)=0
∴ x= −1 orx= −1±
√
5 2
The only acceptable solution isx= −1+
√
5
2 (x≈0.62) as the other two possibilities give negative values forxand we are only considering positive values forx.
Exercise
19C
1 On which of the following curves is d y
d x positive for all values ofx?
a y
x
0
b y
x
0
c y
d y
x
0
e y
x
0
2 On which of the following curves is d y
d x negative for all values ofx?
a y
x 0
b y
x 0
c y
x 0
d
x y
0
e y
x 0
f y
x 0
3 For the function f(x)=2(x−1)2find the values ofxfor which: a f(x)=0 b f(x)=0 c f(x)>0 d f(x)<0 e f(x)= −2
4 For the graph ofy=h(x) shown here find:
Example15 y
x
0 –3, 5 1
2 (4, 6)
1 2
, –3
5 Which of the graphs labelledA–Fcorrespond to each of the graphs labelleda–f? y x 0 y x 0 y x 0 y x 0 y x 0 y x 0 dy dx x 0 dy dx x 0 x 0 dy dx x 0 dy dx x 0 dy dx a b c
d A D
E F B C e f dy dx x 0
6 For the graph ofy= f(x) find:
Example15
a {x: f(x)>0} b {x: f(x)<0} c {x: f(x)=0}
x y
0
(–1, –2)
(1.5, 3)
7 Sketch the graph ofy= f(x) for each of the following
Example16 x y 0 (3, 4) 1 5
a b c d
x y 1 1 0 x y 0 (3, 4) (–1, –3)
y = f(x)
x y
y = f(x) 3
0
8 Find the coordinates of the points on the curvey=x2−5x+6 at which the tangent:
Example17
a makes an angle of 45◦ with the positive direction of thex-axis, i.e. where the gradient is 1
9 Find the coordinates of the points on the parabolay=x2−x−6 at which:
Example17
a the gradient is zero
b the tangent is parallel to the linex+y=6.
10 Use a calculator to plot the graph ofy= f (x)=where: a f(x)=sinx b f(x)=cosx c f(x)=2x
11 A car moves away from a set of traffic lights so that the distanceS(t) metres covered after tseconds is modelled byS(t)=(0.2)t3.
a Find its speed aftertseconds. b What will its speed be whent=1, 3, 5?
12 A rocket is launched from Cape York Peninsula so that aftertseconds its height
h(t) metres is given byh(t)=20t2, 0≤t≤150. After 21
2 minutes this model is no longer appropriate.
a Find the height and the speed of the rocket whent=150. b After how long will its speed be 1000 m/s?
13 The curve with equationy=ax2+bxhas a gradient of 3 at the point (2,−2). a Find the values ofaandb.
b Find the coordinates of the point where the gradient is 0.
19.4
Limits and continuity
Limits
We consider the limit of a function f(x) to be the value that f(x) approaches asxapproaches a given value. lim
x→a f(x)= pmeans that, asxapproachesa,f(x) approachesp. It is important to understand that it is possible to get as close as desired topasxapproachesa. Note that f(x) may or may not be defined atx=a.
For many functionsf(a) is defined, so to evaluate the limit we simply substitute the valuea into the rule for the function.
Example 18
If f(x)=3x2find lim
x→23x 2.
Solution
Since f(x)=3x2is defined atx=2 lim
x→23x
2 =3(2)2
=12
Example 19
For f(x)= 2x
2−5x+2
x−2 ,x =2,find limx→2 f(x).
Solution
Observe that f(x) is defined forx∈R\{2}.
Examine the behaviour of f(x) for values ofx‘near’ 2.
x<2 x>2
f(1.7)=2.4 f(2.3)=3.6 f(1.8)=2.6 f(2.2)=3.4 f(1.9)=2.8 f(2.1)=3.2 f(1.99)=2.98 f(2.01)=3.02 f(1.999)=2.998 f(2.001)=3.002
y
f
x 0
3
1
–1
2 From the table it is apparent that asxtakes values
closer and closer to 2, regardless of whether xapproaches 2 from the left or from the right, the values of f(x) become closer and closer to 3. i.e. lim
x→2 f(x)=3
This may also be seen by observing that:
f(x)= (2x−1)(x−2)
x−2 ,n=2
=2x−1,x =2,
The graph off:R\{2}→R,f(x)=2x−1 is shown.
The following important results are useful for the evaluation of limits: lim
x→c(f(x)+g(x))=xlim→c f(x)+xlim→cg(x) i.e. the limit of the sum is the sum of the limits.
lim
x→ck f(x)=kxlim→c f(x),kbeing a given number lim
x→c(f(x)g(x))=xlim→c f(x) limx→cg(x)
i.e. the limit of the product is the product of the limits.
lim x→c
f(x)
g(x) =
lim x→c f(x)
lim x→cg(x)
,provided lim
x→cg(x)=0
Example 20
Find: a lim
x→0(x
2+2) b lim
x→3
x2−3x
x−3 c xlim→2
x2−x−2 x−2
d lim
x→3(2x+1)(3x−2) e xlim→3
x2−7x+10 x2−25
Solution
a lim x→0(x
2+2)= lim x→0x
2+lim
x→02=0+2=2 b lim
x→3
x2−3x
x−3 =xlim→3
x(x−3)
x−3 =xlim→3x =3
c lim x→2
x2−x−2
x−2 =xlim→2
(x−2)(x+1)
x−2 =xlim→2(x+1)=3
d lim
x→3(2x+1)(3x−2)=xlim→3(2x+1) limx→3(3x−2)=7×7=49 e lim
x→3
x2−7x+10
x2−25 =xlim→3
(x−2)(x−5) (x+5)(x−5) =
lim x→3(x−2) lim x→3(x+5)
= 1
8
Example 21
Find: a lim
h→0(3h+4) b xlim→24x(x+2) c xlim→3
5x+2
x−2
Solution
a lim
h→0(3h+4)=hlim→0(3h)+hlim→0(4)
=0+4
=4
b lim
x→24x(x+2)=xlim→2(4x) limx→2(x+2)
=8×4
=32
c lim x→3
5x+2
x−2 =xlim→3(5x+2)÷xlim→3(x−2)
=17÷1
=17
The notation of limits is used to describe the behaviour of graphs, and a similar notation has been used previously in the book.
Consider f:R\ {0} → R, f(x)= 1
x2.Observe that asx→0, both from the left and from the right, f(x) increases without bound. The limit notation for this is lim
Forg:R\ {0} → R,g(x)= 1
x,the behaviour ofg(x) asxapproaches 0 from the left is different from the behaviour asxapproaches 0 from the right.
With limit notation this is written as:
lim
x→0−g(x)= −∞ and xlim→0+ f(x)= ∞
x y
0
g(x)= 1x
Now examine this function as the magnitude ofxbecomes very large. It can be seen that asxincreases without
bound through positive values, the corresponding values ofg(x) approach zero. Likewise asxdecreases
without bound through negative values, the corresponding values ofg(x) also approach zero.
Symbolically this is written as:
lim
x→∞g(x)=0 and x→−∞lim g(x)=0
Many functions approach a limiting value or limit asxapproaches±∞.
Left and right limits
An idea which is useful in the following discussion is the existence of limits from the left and from the right.
If the value of f(x) approaches the numberpasxapproachesafrom the right-hand side, then it is written as lim
x→a+
f(x)= p.If the value of f(x) approaches the numberpasx approachesafrom the left-hand side, then it is written as lim
x→a−
f(x)= p.
The limit asxapproachesaexists only if the limits from the left and the right both exist and are equal. Then lim
x→a f(x)= p.
The following is an example of when the limit does not exist for a particular value.
Letf(x)=
x3 for 0≤x <1 5 forx=1 6 for 1<x ≤2
It is clear from the graph offthat the lim
x→1 f(x) does not exist. However, ifxis allowed to approach 1 from the left, then
f(x) approaches 1. On the other hand ifxis allowed to approach 1 from the right, then
f(x) approaches 6. Also note that f(1)=5.
x y
0 1
1 2 3 4 5 6
Continuity at a point: informal definition
A function with rule f(x) is said to be continuous whenx=aif the graph ofy= f(x) can be drawn through the point with coordinates (a, f(a)) without a break. Otherwise there is said to be a discontinuity atx=a.
Most of the functions considered in this course are continuous for their domains. A more formal definition of continuity follows.
A functionfis continuous at a pointaif f(a), lim x→a+
f(x) and lim x→a−
f(x) all exist and
are equal. Or equivalently:
A functionfis continuous at the pointx=aif the following three conditions are met: 1 f(x) is defined atx=a 2 lim
x→a f(x) exists 3 xlim→a f(x)= f(a) The function is said to be discontinuous at a point if it is not continuous at that point. A function is said to be continuous everywhere if it is continuous for all real numbers.
The polynomial functions are all continuous forR. The function with rule f(x)= 1 x has a discontinuity atx=0, as f(0) is not defined. It is continuous elsewhere in its domain.
Hybrid functions, as introduced in Chapter 6, provide examples of functions which have points of discontinuity where the function is defined.
Example 22
State the values forxfor which the functions whose graphs are shown below have a
discontinuity. a
x y
0 1
1 2 3
–1
b
x y
0 1
–1 2
1 3
c
x y
0 1 1 2 3
–1
Solution
a There is a discontinuity atx=1, as f(1)=3 but lim x→1+
f(x)= lim
x→1−
f(x)=2.
b There is a discontinuity atx= −1, asf(−1)=2 and lim x→−1−
f(x)=2 but lim
x→−1+
f(x)= −∞and a discontinuity atx=1 as f(1)=2 and lim x→1−
f(x)=2
but lim x→1+
f(x)=3.
c There is a discontinuity atx=1, as f(1)=1 and lim x→1−
f(x)=1 but lim
x→1+
Example 23
For each of the following functions state the values ofxfor which there is a discontinuity and
use the definition of continuity in terms of f(a), lim
x→a+
f(x) and lim
x→a−
f(x) to explain why each
is discontinuous:
a f(x)=
2x ifx ≥0
−2x+1 ifx <0 b f(x)=
x2 ifx≥0
−2x+1 ifx<0
c f(x)=
x ifx≤ −1
x2 if −1<x <0
−2x+1 ifx≥0
d f(x)=
x2+1 ifx≥0
−2x+1 ifx<0
e f(x)=
x ifx ≥0
−2x ifx <0
Solution
a f(0)=0 but lim x→0−
f(x)=1, therefore there is a discontinuity atx=0
b f(0)=0 but lim x→0−
f(x)=1, therefore there is a discontinuity atx=0
c f(−1)= −1 but lim
x→−1+ f(x)=1, therefore there is a discontinuity atx=1 f(0)=1 but lim
x→0− f(x)=0, therefore there is a discontinuity atx=0 d No discontinuity e No discontinuity
Exercise
19D
1 Find the following limits:
Examples18–21
a lim
x→315 b xlim→6(x−5) c lim x→12
(3x−5)
d lim t→−3
(t−2)
(t+5) e tlim→−1
t2+2t +1
t+1 f xlim→0
(x+2)2−4 x
g lim t→1
t2−1
t −1 h xlim→9
√
x+3 i lim
x→0
x2−2x
x
j lim x→2
x3−8
x−2 k xlim→2
3x2−x−10
x2+5x−14 l xlim→1
x2−3x+2
x2−6x+5
2 For each of the following graphs give the values ofxat which a discontinuity occurs. Give
Example22
reasons.
a y
x 2
1 3 4
3 For each of the following functions state the values ofxat which there is a discontinuity
Example23
and use the definition of continuity in terms of f(a), lim x→a+
f(x) and lim x→a−
f(x) to explain
why each stated value ofxcorresponds to a discontinuity.
a f(x)=
3x ifx ≥0
−2x+2 ifx <0 b f(x)=
x2+2 ifx ≥1
−2x+1 ifx <1
c f(x)=
−x ifx ≤ −1
x2 if −1<x <0
−3x+1 ifx ≥0
4 The rule of a particular function is given below. For what values ofxis the graph of this function discontinuous?
y=
2, x <1 (x−4)2−9,1≤x <7
x−7, x ≥7
19.5
When is a function differentiable?
A functionfis said to be differentiable atxif lim h→0
f(x+h)− f(x)
h exists.
The polynomial functions considered in this chapter are differentiable for allx. However this is not true for all functions.
Letf:R→ R, f(x)=
x ifx≥0
−x ifx<0
This function is called the modulus function or absolute value function and it is denoted by f(x)= |x|,f is not differentiable atx=0:
f(0+h)− f(0)
h =
h
h h>0
−h
h h<0
=
1 h>0
−1 h<0
y = x y
x 0
So lim h→0
f(0+h)− f(0)
h does not exist. i.e.fis not differentiable atx=0.
Example 24
Letf:R→R, f(x)=
x ifx ≥0
−x ifx <0 (= |x|) Sketch the graph of the derivative for a suitable domain.
Solution
f(x)=
1 ifx >0
−1 ifx<0 f(x) is not defined atx=0
y
x 1
–1 0
Example 25
Draw a sketch graph of fwhere the graph offis as illustrated.
Indicate where fis not defined.
y
x 1 2
0 –1
Solution
The derivative does not exist atx=0; i.e. the function is not differentiable atx=0.
x y
2
1 0 –1
–2
It was shown in the previous section that some hybrid functions are continuous forR. There are hybrid functions which are differentiable forR. The smoothness of the ‘joins’ determines if this is the case.
Example 26
For the function with following rule find f(x) and sketch the graph ofy= f(x):
f(x)=
x2+2x+1 ifx≥0
2x+1 ifx<0
Solution
f(x)=
2x+2 ifx ≥0 2 ifx <0 In particular f(0) is defined and is equal to 2. Also f(0)=1. The two sections of the graph ofy= f(x) join smoothly at (0, 1).
2
Example 27
For the function with rule
f(x)=
x2+2x+1 ifx ≥0 x+1 ifx<0
state the set of values for which the derivative is defined, find f(x) for this set of values and
sketch the graph ofy= f(x).
Solution
f(x)=
2x+2 ifx>0 1 ifx<0
f(0) is not defined as the limits from the left and right are not equal. The function is differentiable forR\{0}.
1 2
0 x
y
Exercise
19E
1 In each of the figures below a function graphfis given. Sketch the graph of f . Obviously
Examples24–26
your sketch of fcannot be exact, but f(x) should be 0 at values ofxfor which the gradient offis zero; f(x) should be<0 where the original graph slopes downward, and so on.
a
x f y
1 0 –1
b
x f
y
4 2 –2 –4 0
c
x f y
1 0 –1
d
x f y
1 0 –1
e
x f y
1 0 –1
f
x f y
1 0 –1
2 For the function with following rule find f (x) and sketch the graph ofy= f (x):
Example26
f(x)=
3 For the function with following rule state the set of values for which the derivative is
Example27
defined, find f(x) for this set of values and sketch the graph ofy= f(x):
f(x)=
x2+2x+1 ifx ≥1
−2x+3 ifx <1
4 For the function with following rule state the set of values for which the derivative is
Example27
defined, find f(x) for this set of values and sketch the graph ofy= f(x):
f(x)=
−x2−3x+1 ifx ≥ −1
Review
Chapter summary
The notation for limit ashapproaches 0 is written as lim h→0. The following are important results for limits:
r lim
x→c(f(x)+g(x))=limx→c f(x)+xlim→cg(x) i.e. the limit of the sum is the sum of the limits.
r lim
x→ck f(x)=kxlim→c f(x),kbeing a given number
r lim
x→c(f(x)g(x))=xlim→c f(x) limx→cg(x)
i.e. the limit of the product is the product of the limits.
r lim x→c
f(x)
g(x) =
lim x→c f(x)
lim x→cg(x)
,provided lim
x→cg(x)=0
i.e. the limit of the quotient is the quotient of the limits.
A functionfis defined as continuous at the pointx=aif three conditions are met: 1 f(x) is defined atx=a
2 lim
x→af(x) exists 3 lim
x→af(x)= f(a)
The function is said to be discontinuous at a point if it is not continuous at that point. We say that a function is continuous everywhere if it is continuous for all real numbers. For the graph ofy= f(x) of the functionf:R→R:
The gradient of the chordPQ= f(x+h)− f(x)
h .
The gradient of the graph atPis given by
lim h→0
f(x+h)− f(x)
h .
This limit gives a rule for the derived function denoted by f, where f:R→Rand
f (x)= lim
h→0
f(x+h)− f(x)
h
x y
0
P (x, f(x))
Q (x + h, f(x + h))
The general rule of the derived function of f(x)=xn,n=1, 2, 3, . . . For f(x)=xn, fx=nxn−1,n=1,2,3, . . .
For f(x)=1, f(x)=0 For example:
For f(x)=x2, f(x)=2x. For f(x)=x3, f (x)=3x2. For f(x)=x4, f(x)=4x3. For f(x)=1, f(x)=0. The derivative of a number multiple is the multiple of the derivative:
Forg(x)=k f(x), wherekis a constant
g(x)=k f(x)
Review
The derivative of a constant is always zero: Forg(x)=a,g(x)=0
For example: f(x)=27.3, f(x)=0
The derivative of the sum is the sum of the derivatives:
If f(x)=g(x)+h(x) then
f (x)=g(x)+h(x).
For example: f(x)=x2+x3, f(x)=2x+3x2
g(x)=3x2+43,g(x)=3(2x)+4(3x2)=6x+12x2 At a point (a,g(a)) on the curvey=g(x) the gradient isg(a).
Forx<bthe gradient is positive, i.e.g(x)>0 x=bthe gradient is zero, i.e.g(b)=0
b<x<athe gradient is negative, i.e.g(a)<0 x=athe gradient is zero, i.e.g(a)=0
x>athe gradient is positive, i.e.g(x)>0
y
x a
b
y = g(x) R(b, g(b))
S(a, g(a)) 0
Multiple-choice questions
1 The gradient of the curvey=x3+4xat the point wherex=2 is
A 12 B 4 C 10 D 16 E 8
2 The gradient of the chord of the curvey=2x2between the points wherex=1 and x=1+his given by
A 2(x+h)2−2x2 B 4+2h C 4 D 4x E 4+h 3 Ify =2x4−5x3+2, then d y
d x equals
A 8x3−5x2+2 B 4x4−15x2+2 C 4x4−10x2 D 8x3−15x+2 E 8x3−15x2
4 If f(x)=x2(x+1) then f(−1) equals
A −1 B 1 C 2 D −2 E 5
5 If f(x)=(x−3)2, then f (x) equals
A x−3 B x−6 C 2x−6 D 2x+9 E 2x
6 Ify= 2x 4+9x2
3x , then d y d x equals
A 2x 4
3 +6x B 2x+3 C 2x
2+3 D 8x
3+18x
3 E 8x
3+18x 7 Given thaty=x2−6x+9, the values ofxfor which d y
d x ≥0 are
A x ≥3 B x >3 C x ≥ −3 D x ≤ −3 E x <3 8 Ify=2x4−36x2, the points at which the tangent to the curve is parallel to thex-axis are
Review
9 The coordinates of the point on the graph ofy=x2+6x−5 at which the tangent is parallel to the liney=4x are
A (−1,−10) B (−1,−2) C (1, 2) D (−1, 4) E (−1, 10) 10 Ify= −2x3+3x2−x+1, then d y
d x is equals
A 6x2+6x−1 B −6x2+6x C −6x2+3x−1 D −6x2+6x−1 E 6x2−6x−1
Short-answer questions (technology-free)
1 Find d y d x when:
a y=3x2−2x+6 b y=5 c y=2x(2−x) d y=4(2x−1)(5x+2) e y=(x+1)(3x−2) f y=(x+1)(2−3x)
2 Find d y d x when:
a y= −x b y=10 c y= (x+3)(2x+1)
4
d y= 2x 3−x2
3x e y=
x4+3x2
2x2
3 For each of the following functions find they-coordinates and the gradient at the point on the curve for the given value ofx:
a y=x2−2x+1,x=2 b y=x2−2x, x = −1 c y=(x+2)(x−4),x =3 d y=3x2−2x3, x = −2
4 Find the coordinates of the points on the curves given by the following equations at which the gradient has the given value:
a y=x2−3x+1; d y
d x =0 b y=x
3−6x2+4; d y
d x = −12
c y=x2−x3; d y
d x = −1 d y=x
3−2x+7; d y
d x =1
e y=x4−2x3+1; d y
d x =0 f y=x(x−3)
2 ; d y
d x =0
5 For the function with rule f(x)=3(2x−1)2find the values ofxfor which: a f(x)=0 b f(x)=0 c f(x)>0 d f(x)<0 e f(x)>0 f f(x)=3 6 The curve with equationy=ax2+bxhas a gradient of 3 at the point (1, 1). Find:
a the values ofaandb b the coordinates of the points where the gradient is 0.
7 Sketch the graph ofy= f(x). (All details cannot be determined but the axis intercepts and shape of graph can be determined.)
y
x
2 5
–1
Review
8 For the graph ofy=h(x) find: a {x:h(x)>0}
b {x:h(x)<0} c {x:h(x)=0}
y
y = h(x)
x
0
(–1, –4) (4, 3)
Extended-response questions
1 The diagram to the right shows part of the graph
of d y
d x againstx.
x dx
dy
–1 0 2 5
Sketch a possible shape ofyagainstxover the same interval if:
r y= −1 whenx= −1
r y=0 whenx=0
r y=1 whenx=2.
2 The graph shown is that of a polynomial of the form
P(x)=ax3+bx2+cx+d.
Find the values ofa,b,candd. Note:Q(1,−2) is not a turning point.
y
x
Q(1, –2) R(–2, 3)
3 A body moves in a path described by the equationy= 1 5x
5+1 2x
4,x ≥0.
Units are in kilometres andxandyare the horizontal and vertical axes respectively. a What will be the direction of motion (give the answer as angle between direction of
motion and thex-axis) when thex-value is i 1 km? ii 3 km?
b Find a value ofxfor which the gradient of the path is 32.
4 A trail over a mountain pass can be modelled by the curve with equation
y=2+0.12x−0.01x3, wherexandyare, respectively, the horizontal and vertical distances measured in kilometres, 0≤x≤3.
a Find the gradients at the beginning and the end of the trail.
b Calculate the point where the gradient is zero, and calculate also the height of the pass.
5 A tadpole begins to swim vertically upwards in a pond and aftertseconds it is 25−0.1t3cm below the surface.
a How long does the tadpole take to reach the surface, and what is its speed then? b What is the average speed over this time?
6 a Show that the gradients of the curvey =x(x−2) at the points (0, 0) and (2, 0) only differ in sign. What is the geometrical interpretation for this?
b If the gradient of the curvey=x(x−2)(x−5) at the points (0, 0), (2, 0) and (5, 0) are l,mandnrespectively, show that 1
l +
1
m +
1
