MATH 205, Chapter 2. A function is a rule that assigns each number in one set A to exactly one number in a second set B

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MATH 205, Chapter 2

Introductory Calculus

James Madison University

Introductory Calculus MATH 205, Chapter 2

A function is a rule that assigns each number in one set A to exactly one number in a second set B

Set A = Domain of the function.

Range = set consisting of elements in set B that are associated with an element of A

Example - let f be the function that assigns each real number to its square.

f : x→ x² or f (x) = x²

f - “the function”

f (x) the number that the function assigns to some input value x.

f (x) = x²

Domain of f ? Range of f ?

Find I f (3) I f (w− 2) I _{f (w}^{f (z)}3+1)+2

I f (x+h)−f (x) h

Suppose:

I g (x) =√ x²+ 2x I h(x) = _x2^x+1²

I k(x) =_9x^2x2−4

Domain and Range for g , h, k?

Suppose:

I g (x) =√ x²+ 2x Domain and Range for g ?

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Suppose:

I h(x) = _x2^x+1²

Domain and Range for h?

Suppose:

I k(x) =_9x^2x2−4

Domain and Range for k?

What are the domain and range of f ? g ?

Suppose:

I f (x) = x² I g (x) =√

x²+ 2x I h(x) =_x2^x+1²

I k(x) =_9x^2x2−4

Find

I (f + g )(x) I _g^h(x) =^h(x)_{g (x)} I k◦ f (x) = k(f (x)) I fg (x) (or (f∗ g)(x)) I g◦ f (x) = g(f (x)) I f ◦ g(x) = f (g(x))

Find

I (f + g )(2) I ^f_g(2) I g◦ f (3) I fg (2)

I f ◦ g(2) (approx.)

Rental charges R for driving a rental car x miles are given by the formula:

R(x) =

30 if 0≤ x ≤ 100

30 + 0.32(x− 100) if x > 100

Describe the rental pricing policy in words.

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A box is to be constructed by cutting an x by x square from each corner of a 24 by 36 rectangular piece of cardboard and folding up the sides. If function V is the volume of the resulting box, write a formula for V (x). What is the domain of this function? More difficult question: what is the range?

Let f (x) =√ ^x−2 x²+6x−4.

What is f (0)? What is f (1)? What is f (2)?

Sketch a graph. (calculator?)

What happens to the values for f (x) when x is very close to but not equal to 2?

By restricting x to values sufficiently close to (but not quite equal to) 2, we can make the values of f (x) stay as close as we want to a particular number = ...

We say that the limit of f as x approaches 2 is this number, and write

x→2limf (x) = lim

x→2

x− 2

√x²+ 6x− 4= ...

Some similar limit questions:

I lim_x→2^x_x−2²⁻⁴ I lim_x→2^√_x−3^x²⁺⁴ I lim_x→1x³+ x− 3 I lim_x→2^x_x−2²⁺⁴

The first is similar to lim_x→2√ ^x−2

x²+6x−4, except that it should be much easier to find a value for the limit without a calculator.The 2nd and 3rd are a bit different, but rather easier questions. The last has problems all its own.

lim_x→2^x_x−2²⁻⁴and lim_x→2^√ ^x−2

x²+6x−4 are more interesting limits than lim_x→2^√_x−3^x²⁺⁴. The first two are sometimes called “indeterminate forms.” For the first two limits, both the numerator and denominator are getting close to 0. “ ⁰₀ ” is not a number ... it really is not anything at all other than a meaningless expression.

We have to do some additional work to figure out what the limit is.

In lim_x→2^√_x−3^x²⁺⁴, the numerator is getting close to√

4 + 4 = 2√ 2 and the denominator is getting close to 2-3 = -1, so the whole fraction has limit ²₋₁^√² =−2√

2.

A formal definition of the idea of limit is a bit difficult. Repeating the somewhat oversimplified version:

We say that the limit of function f as x approaches point a is equal to some number L, in symbols

x→alimf (x) = L

if we can guarantee that the values of f (x) are as close as we might wish to the number L by restricting x values to those sufficiently close to (but not equal to) the point a.

Note: The “limit” does not in any way depend on the actual value of f at the point a. f (a) may not even exist, or f (a) may be a number that is different from L. It is true that often f (a) = L as it did in some of the previous examples, but this is in no way required in order for the limit to exist.

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Another example:

x→0lim

p2^x²− 1 x

Explain why this limit does not exist. A calculator may be a help.

A similar example for which a calculator may not be needed:

x→2lim

|x − 2|

x− 2

In both of the above, when we look for a limit we get different answers depending on the “direction” we are coming from.

In the case of

√2^x2−1

x , we could say that if we approach 0 using only numbers greater than 0, i.e. “from the right,” then there is a limit, and we can write

x→0lim⁺

p2^x²− 1

x ≈ 0.83255 (the actual number isp

ln(2) ). Similarly we have:

x→0lim⁻

p2^x²− 1

x ≈ −0.83255

x→2lim⁺

|x − 2|

x− 2 = 1

x→2lim⁻

|x − 2|

x− 2 =−1 Note again that lim_x→0

√2^x2−1

x and lim_x→2^|x−2|_x−2 do not exist.

Overall there is a limit if both the left and right limit exist and the left and right limit are the same number.

In the above, what if anything can you say about the limits as x approaches -2 or as x approaches 0 or as x approaches +2?

Overall? from the left? from the right?

Some more examples:

zlim→3

z²− 2z − 3

z²− 9 lim

z→1

z²− 2z − 3 z²− 9

zlim→3

z²− 2z + 3

z²− 9 lim

w→2

w− 2 w²− 4w + 4

h→0lim

(3 + h)²+ (3 + h)− 12

h lim

h→0(1 + h)¹^h

We also on occasion will discuss limits “at infinity” (at∞). As is usually the case with the symbol∞ we are being a little loose in how we speak (so to speak). When we ask what happens to a function “as x→ ∞” we are really just asking what happens as x gets arbitrarily large. x→ −∞ means x gets arbitrarily large in the negative direction.

For example, it should not be hard to interpret the meaning of

x→∞lim 1 + x 1− x

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Limits as x→ ±∞ are always directly related to the existence of horizontal asymptotes. Some additional examples are below. In each case, you might be able to reason your way to an answer without a graph, but a graph would certainly settle the matter.

zlim→∞

z²− 2z − 3

z²− 9 lim

z→∞

3z²− 2z − 3 5z²− 9

z→∞lim

z³− 2z − 3

z− 9 lim

z→∞

z²− 2z − 3 z⁵− 9

z→∞lim

z³− 2z − 3

2^z lim

z→∞

pz²+ z− z

Introductory Calculus MATH 205, Chapter 2 Introductory Calculus MATH 205, Chapter 2

The notion of a continuous function or of continuity is fairly intuitive. Assuming that below is the graph of y = g (x) for some function g , we would say that g is continuous everywhere except 4 values of x. Where is g discontinuous?

Formal definition in symbols:

Function f is continuous at point a if a is in the domain of f (i.e.

if f (a) actually exists), the limit lim_x→af (x) exists, and

x→alimf (x) = f (a)

Only one of the following functions is continuous at x = 2. Which one is continuous at 2, and why do the others fail to be

continuous?

f (x) =

( 3 if x = 2

x²−4

x−2 otherwise g (x) =^x_x−2²⁻⁴

h(x) =

( 4 if x = 2

x²−4

x−2 otherwise k(x) =

( 8 if x = 2

x²+4

x−2 otherwise

For an object in motion, we will talk about the following quantities. t represents time.

I s(t) = the position of the object at time t, i.e. the distance from some identified point. Typical units are ft, miles, km, etc.

I v (t) = the velocity of the object at time t, i.e. the rate at which the position is changing. Typical units are ft/sec, miles/hr, km/min, miles/sec etc.

I a(t) = the acceleration of the object at time t, i.e. the rate at which the velocity is changing. Typical units are ft/sec per sec (ft/sec²), miles/hr per hour (mi/hr²), km/min per minute (km/min²), etc.

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Imagine the following: A bicycle is initially not moving, but beginning to move at time t = 0 (so that velocity v (0) = 0 but v (t) > 0 when t > 0). Also let s(0) = 0 (recalling that we can define the “coordinates” for our starting position anyway we want).

If the bicycle has constant acceleration, here is a possible formula for position s in feet after t seconds (later: how did we get this??).

s(t) =3 2t².

What is the position after 4 seconds (what is s(4))?

What is s(6)?

What is the average velocity of the bicycle between t = 4 and t = 6?

In other words, starting at t = 4, the average velocity over the next 2 seconds is ...?

Starting at t = 4 what is the average velocity ...

I ... over the next 1 second?

I ... 0.5 second?

I ... 0.1 second?

I ... 0.01 second?

I ... over the next h seconds? (Where h is an arbitrary but presumably small number.)

We should be able to guess the exact value for the velocity at four seconds.

It also seems reasonable (I hope!?) to say the following:

The average velocity between time 4 and time 4+h is ...

s(4 + h)− s(4)

h =

3

2(4 + h)²− 24 h

=

32(16 + 8h + h²)− 24 h

= 12h +³₂h² h

= 12 +3 2h

Velocity is the rate of change of position. Average velocity over an interval (of time) is the average rate of change of position.

The exact velocity (or exact rate of change of position) at time 4 is the limit of the average rate of change between time 4 and time 4+h as h gets very small. Or in symbols:

v (4) = lim

h→0

s(4 + h)− s(4)

h .

If t is any particular time (t could be 2 or 3 or 1.67 or whatever) then the velocity at time t is

v (t) = lim

h→0

s(t + h)− s(t)

h .

The rate of change of s (or any function, not necessarily

representing position) is also called the derivative of s and denoted s⁰. So we could write

s⁰(t) = lim

h→0

s(t + h)− s(t)

h .

There is nothing special about position functions. If f is any function, then the derivative f⁰(or rate of change) of f is given by

f⁰(x) = lim

h→0

f (x + h)− f (x)

h .

Note that there is no guarantee that the limit above actually exists. Sometimes limits exists, sometimes not. When and where the limit does exist (and thus the derivative exists) we say that the function is differentiable.

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The derivative graphically:

The average rate of change between point x₀and point x₀+ h is the slope of the line joining two points (x₀, f (x₀)) and

(x0+ h, f (x0+ h)). This is called a secant line.

The actual (instantaneous) rate of change at a point is the slope of the line tangent to the curve at that point. The slope of the tangent line is given by f⁰, the limit of the slopes of the secant lines:

f⁰(x0) = lim

h→0

f (x0+ h)− f (x0)

h .