Contributed by:

In this section, we will learn about:

The interaction between calculus and calculators. We start with a graph produced by a graphing calculator or computer, and then we refine it. We use calculus to make sure that we reveal all the important aspects of the curve.

The interaction between calculus and calculators. We start with a graph produced by a graphing calculator or computer, and then we refine it. We use calculus to make sure that we reveal all the important aspects of the curve.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

The method we used to sketch curves

in Section 4.5 was a culmination of much

of our study of differential calculus.

The graph was the final object we produced.

The method we used to sketch curves

in Section 4.5 was a culmination of much

of our study of differential calculus.

The graph was the final object we produced.

3.
APPLICATIONS OF DIFFERENTIATION

In this section, our point of view is

completely different.

We start with a graph produced by a graphing

calculator or computer, and then we refine it.

We use calculus to make sure that we reveal

all the important aspects of the curve.

In this section, our point of view is

completely different.

We start with a graph produced by a graphing

calculator or computer, and then we refine it.

We use calculus to make sure that we reveal

all the important aspects of the curve.

4.
APPLICATIONS OF DIFFERENTIATION

With the use of graphing devices,

we can tackle curves that would be

far too complicated to consider without

With the use of graphing devices,

we can tackle curves that would be

far too complicated to consider without

5.
APPLICATIONS OF DIFFERENTIATION

4.6

Graphing with

Calculus and Calculators

In this section, we will learn about:

The interaction between calculus and calculators.

4.6

Graphing with

Calculus and Calculators

In this section, we will learn about:

The interaction between calculus and calculators.

6.
CALCULUS AND CALCULATORS Example 1

Graph the polynomial

f(x) = 2x6 + 3x5 + 3x3 – 2x2

Use the graphs of f’ and f” to estimate all

maximum and minimum points and intervals

of concavity.

Graph the polynomial

f(x) = 2x6 + 3x5 + 3x3 – 2x2

Use the graphs of f’ and f” to estimate all

maximum and minimum points and intervals

of concavity.

7.
CALCULUS AND CALCULATORS Example 1

If we specify a domain but not

a range, many graphing devices will

deduce a suitable range from the values

If we specify a domain but not

a range, many graphing devices will

deduce a suitable range from the values

8.
CALCULUS AND CALCULATORS Example 1

The figure shows the plot from one such

device if we specify that -5 ≤ x ≤ 5.

The figure shows the plot from one such

device if we specify that -5 ≤ x ≤ 5.

9.
CALCULUS AND CALCULATORS Example 1

This viewing rectangle is useful for showing

that the asymptotic behavior (or end behavior)

is the same as for y = 2x6.

However, it is

obviously hiding

some finer detail.

This viewing rectangle is useful for showing

that the asymptotic behavior (or end behavior)

is the same as for y = 2x6.

However, it is

obviously hiding

some finer detail.

10.
CALCULUS AND CALCULATORS Example 1

So, we change to the viewing rectangle

[-3, 2] by [-50, 100] shown here.

So, we change to the viewing rectangle

[-3, 2] by [-50, 100] shown here.

11.
CALCULUS AND CALCULATORS Example 1

From this graph, it appears:

There is an absolute minimum value of about -15.33

when x ≈ -1.62 (by using the cursor).

f is decreasing

on (-∞, -1.62)

and increasing

on (-1.62, -∞).

From this graph, it appears:

There is an absolute minimum value of about -15.33

when x ≈ -1.62 (by using the cursor).

f is decreasing

on (-∞, -1.62)

and increasing

on (-1.62, -∞).

12.
CALCULUS AND CALCULATORS Example 1

Also, it appears:

There is a horizontal tangent at the origin and

inflection points when x = 0 and when x is somewhere

between -2 and -1.

Also, it appears:

There is a horizontal tangent at the origin and

inflection points when x = 0 and when x is somewhere

between -2 and -1.

13.
CALCULUS AND CALCULATORS Example 1

Now, let’s try to confirm these impressions

using calculus.

We differentiate and get:

f’(x) = 12x5 + 15x4 + 9x2 – 4x

f”(x) = 60x4 + 60x3 + 18x – 4

Now, let’s try to confirm these impressions

using calculus.

We differentiate and get:

f’(x) = 12x5 + 15x4 + 9x2 – 4x

f”(x) = 60x4 + 60x3 + 18x – 4

14.
CALCULUS AND CALCULATORS Example 1

When we graph f’ as in this figure, we see

that f’(x) changes from negative to positive

when x ≈ -1.62

This confirms—by

the First Derivative

Test—the minimum

value that we found

earlier.

When we graph f’ as in this figure, we see

that f’(x) changes from negative to positive

when x ≈ -1.62

This confirms—by

the First Derivative

Test—the minimum

value that we found

earlier.

15.
CALCULUS AND CALCULATORS Example 1

However, to our surprise, we also notice

that f’(x) changes from positive to negative

when x = 0 and from negative to positive

when x ≈ 0.35

However, to our surprise, we also notice

that f’(x) changes from positive to negative

when x = 0 and from negative to positive

when x ≈ 0.35

16.
CALCULUS AND CALCULATORS Example 1

This means that f has a local maximum at 0

and a local minimum when x ≈ 0.35, but these

were hidden in the earlier figure.

This means that f has a local maximum at 0

and a local minimum when x ≈ 0.35, but these

were hidden in the earlier figure.

17.
CALCULUS AND CALCULATORS Example 1

Indeed, if we now zoom in toward the

origin, we see what we missed before:

A local maximum value of 0 when x = 0

A local minimum

value of about -0.1

when x ≈ 0.35

Indeed, if we now zoom in toward the

origin, we see what we missed before:

A local maximum value of 0 when x = 0

A local minimum

value of about -0.1

when x ≈ 0.35

18.
CALCULUS AND CALCULATORS Example 1

What about concavity and inflection points?

From these figures, there appear to be

inflection points when x is a little to the left

of -1 and when x is a little to the right of 0.

What about concavity and inflection points?

From these figures, there appear to be

inflection points when x is a little to the left

of -1 and when x is a little to the right of 0.

19.
CALCULUS AND CALCULATORS Example 1

However, it’s difficult to determine

inflection points from the graph of f.

So, we graph the second derivative f” as follows.

However, it’s difficult to determine

inflection points from the graph of f.

So, we graph the second derivative f” as follows.

20.
CALCULUS AND CALCULATORS Example 1

We see that f” changes from positive to

negative when x ≈ -1.23 and from negative

to positive when x ≈ 0.19

We see that f” changes from positive to

negative when x ≈ -1.23 and from negative

to positive when x ≈ 0.19

21.
CALCULUS AND CALCULATORS Example 1

So, correct to two decimal places, f is

concave upward on (-∞, -1.23) and (0.19, ∞)

and concave downward on (-1.23, 0.19).

The inflection points

are (-1.23, -10.18)

and (0.19, -0.05).

So, correct to two decimal places, f is

concave upward on (-∞, -1.23) and (0.19, ∞)

and concave downward on (-1.23, 0.19).

The inflection points

are (-1.23, -10.18)

and (0.19, -0.05).

22.
CALCULUS AND CALCULATORS Example 1

We have discovered that no single

graph shows all the important features

of the polynomial.

We have discovered that no single

graph shows all the important features

of the polynomial.

23.
CALCULUS AND CALCULATORS Example 1

However, when taken together, these two

figures do provide an accurate picture.

However, when taken together, these two

figures do provide an accurate picture.

24.
CALCULUS AND CALCULATORS Example 2

Draw the graph of the function 2

x + 7x + 3

f ( x) = 2

x

in a viewing rectangle that contains

all the important features of the function.

Estimate the maximum and minimum values

and the intervals of concavity.

Then, use calculus to find these quantities exactly.

Draw the graph of the function 2

x + 7x + 3

f ( x) = 2

x

in a viewing rectangle that contains

all the important features of the function.

Estimate the maximum and minimum values

and the intervals of concavity.

Then, use calculus to find these quantities exactly.

25.
CALCULUS AND CALCULATORS Example 2

This figure, produced by a computer

with automatic scaling, is a disaster.

This figure, produced by a computer

with automatic scaling, is a disaster.

26.
CALCULUS AND CALCULATORS Example 2

Some graphing calculators use

[-10, 10] by [-10, 10] as the default

viewing rectangle.

So, let’s try it.

Some graphing calculators use

[-10, 10] by [-10, 10] as the default

viewing rectangle.

So, let’s try it.

27.
CALCULUS AND CALCULATORS Example 2

We get this graph—which is a major

The y-axis appears to be a vertical asymptote.

It is because:

x2 + 7 x + 3

lim 2

=∞

x→ 0 x

We get this graph—which is a major

The y-axis appears to be a vertical asymptote.

It is because:

x2 + 7 x + 3

lim 2

=∞

x→ 0 x

28.
CALCULUS AND CALCULATORS Example 2

The figure also allows us to estimate

the x-intercepts: about -0.5 and -6.5

The exact values are obtained by using

the quadratic formula

to solve the equation

x2 + 7x + 3 = 0

We get:

x = (-7 ± 37 )/2

The figure also allows us to estimate

the x-intercepts: about -0.5 and -6.5

The exact values are obtained by using

the quadratic formula

to solve the equation

x2 + 7x + 3 = 0

We get:

x = (-7 ± 37 )/2

29.
CALCULUS AND CALCULATORS Example 2

To get a better look at horizontal

asymptotes, we change to the viewing

rectangle [-20, 20] by [-5, 10], as follows.

To get a better look at horizontal

asymptotes, we change to the viewing

rectangle [-20, 20] by [-5, 10], as follows.

30.
CALCULUS AND CALCULATORS Example 2

It appears that y = 1 is the horizontal

This is easily confirmed:

x2 + 7 x + 3

lim

x → ±∞ x2

⎛ 7 3 ⎞

= lim ⎜1 + + 2 ⎟

x → ±∞

⎝ x x ⎠

=1

It appears that y = 1 is the horizontal

This is easily confirmed:

x2 + 7 x + 3

lim

x → ±∞ x2

⎛ 7 3 ⎞

= lim ⎜1 + + 2 ⎟

x → ±∞

⎝ x x ⎠

=1

31.
CALCULUS AND CALCULATORS Example 2

To estimate the minimum value,

we zoom in to the viewing rectangle

[-3, 0] by [-4, 2], as follows.

To estimate the minimum value,

we zoom in to the viewing rectangle

[-3, 0] by [-4, 2], as follows.

32.
CALCULUS AND CALCULATORS Example 2

The cursor indicates that the absolute

minimum value is about -3.1 when x ≈ -0.9

We see that the function

decreases on

(-∞, -0.9) and (0, ∞)

and increases on

(-0.9, 0)

The cursor indicates that the absolute

minimum value is about -3.1 when x ≈ -0.9

We see that the function

decreases on

(-∞, -0.9) and (0, ∞)

and increases on

(-0.9, 0)

33.
CALCULUS AND CALCULATORS Example 2

We get the exact values by differentiating:

7 6 7x + 6

f '( x) = − 2 − 3 = − 3

x x x

This shows that f’(x) > 0 when -6/7 < x < 0

and f’(x) < 0 when x < -6/7 and when x > 0.

⎛ 6⎞ 37

The exact minimum value is: f ⎜− ⎟ = − ≈−3.08

⎝ 7⎠ 12

We get the exact values by differentiating:

7 6 7x + 6

f '( x) = − 2 − 3 = − 3

x x x

This shows that f’(x) > 0 when -6/7 < x < 0

and f’(x) < 0 when x < -6/7 and when x > 0.

⎛ 6⎞ 37

The exact minimum value is: f ⎜− ⎟ = − ≈−3.08

⎝ 7⎠ 12

34.
CALCULUS AND CALCULATORS Example 2

The figure also shows that an inflection point

occurs somewhere between x = -1 and x = -2.

We could estimate it much more accurately using

the graph of the second derivative.

In this case, though,

it’s just as easy

to find exact values.

The figure also shows that an inflection point

occurs somewhere between x = -1 and x = -2.

We could estimate it much more accurately using

the graph of the second derivative.

In this case, though,

it’s just as easy

to find exact values.

35.
CALCULUS AND CALCULATORS Example 2

Since 14 18 2(7 x + 9)

f ''( x) = 3 + 4 = 4

x x x

we see that f”(x) > 0 when x > -9/7 (x ≠ 0).

So, f is concave upward on (-9/7, 0) and (0, ∞)

and concave downward on (-∞, -9/7).

The inflection point is: ⎛ 9 71 ⎞

⎜− , − ⎟

⎝ 7 27 ⎠

Since 14 18 2(7 x + 9)

f ''( x) = 3 + 4 = 4

x x x

we see that f”(x) > 0 when x > -9/7 (x ≠ 0).

So, f is concave upward on (-9/7, 0) and (0, ∞)

and concave downward on (-∞, -9/7).

The inflection point is: ⎛ 9 71 ⎞

⎜− , − ⎟

⎝ 7 27 ⎠

36.
CALCULUS AND CALCULATORS Example 2

The analysis using the first two derivatives

shows that these figures display all the major

aspects of the curve.

The analysis using the first two derivatives

shows that these figures display all the major

aspects of the curve.

37.
CALCULUS AND CALCULATORS Example 3

Graph the function 2 3

x ( x + 1)

f ( x) = 2 4

( x −2) ( x −4)

Drawing on our experience

with a rational function

in Example 2, let’s start

by graphing f in the

viewing rectangle

[-10, 10] by [-10, 10].

Graph the function 2 3

x ( x + 1)

f ( x) = 2 4

( x −2) ( x −4)

Drawing on our experience

with a rational function

in Example 2, let’s start

by graphing f in the

viewing rectangle

[-10, 10] by [-10, 10].

38.
CALCULUS AND CALCULATORS Example 3

From the figure, we have the feeling

that we are going to have to zoom in to see

some finer detail and also zoom out to see

the larger picture.

From the figure, we have the feeling

that we are going to have to zoom in to see

some finer detail and also zoom out to see

the larger picture.

39.
CALCULUS AND CALCULATORS Example 3

However, as a guide to intelligent

zooming, let’s first take a close look

at the expression for f(x).

However, as a guide to intelligent

zooming, let’s first take a close look

at the expression for f(x).

40.
CALCULUS AND CALCULATORS Example 3

Due to the factors (x – 2)2 and (x – 4)4 in

the denominator, we expect x = 2 and x = 4

to be the vertical asymptotes.

2 3

x ( x + 1)

Indeed, lim 2 4

=∞

x → 2 ( x −2) ( x −4)

and

2 3

x ( x + 1)

lim 2 4

=∞

x → 4 ( x −2) ( x −4)

Due to the factors (x – 2)2 and (x – 4)4 in

the denominator, we expect x = 2 and x = 4

to be the vertical asymptotes.

2 3

x ( x + 1)

Indeed, lim 2 4

=∞

x → 2 ( x −2) ( x −4)

and

2 3

x ( x + 1)

lim 2 4

=∞

x → 4 ( x −2) ( x −4)

41.
CALCULUS AND CALCULATORS Example 3

To find the horizontal asymptotes, we divide

the numerator and denominator by x6:

3

2

x ( x + 1)3 1⎛ 1⎞

2 3 ⋅ 3 ⎜1+ ⎟

x ( x + 1) x 3

x x⎝ x⎠

2

= 2 4

= 2 4

( x −2) ( x −4) 4

( x − 2) ( x − 4) ⎛ 2⎞ ⎛ 4⎞

2

⋅ 4 ⎜1 − ⎟ ⎜1 − ⎟

x x ⎝ x⎠ ⎝ x⎠

This shows that f(x) → 0 as x → ± ∞.

So, the x-axis is a horizontal asymptote.

To find the horizontal asymptotes, we divide

the numerator and denominator by x6:

3

2

x ( x + 1)3 1⎛ 1⎞

2 3 ⋅ 3 ⎜1+ ⎟

x ( x + 1) x 3

x x⎝ x⎠

2

= 2 4

= 2 4

( x −2) ( x −4) 4

( x − 2) ( x − 4) ⎛ 2⎞ ⎛ 4⎞

2

⋅ 4 ⎜1 − ⎟ ⎜1 − ⎟

x x ⎝ x⎠ ⎝ x⎠

This shows that f(x) → 0 as x → ± ∞.

So, the x-axis is a horizontal asymptote.

42.
CALCULUS AND CALCULATORS Example 3

It is also very useful to consider the behavior

of the graph near the x-intercepts using an

analysis like that in Example 11 in Section 2.6

Since x2 is positive, f(x) does not change sign at 0,

so its graph doesn’t cross the x-axis at 0.

However, due to the factor (x + 1)3, the graph does

cross the x-axis at -1 and has a horizontal tangent there.

It is also very useful to consider the behavior

of the graph near the x-intercepts using an

analysis like that in Example 11 in Section 2.6

Since x2 is positive, f(x) does not change sign at 0,

so its graph doesn’t cross the x-axis at 0.

However, due to the factor (x + 1)3, the graph does

cross the x-axis at -1 and has a horizontal tangent there.

43.
CALCULUS AND CALCULATORS Example 3

Putting all this information together, but

without using derivatives, we see that the

curve has to look something like the one here.

Putting all this information together, but

without using derivatives, we see that the

curve has to look something like the one here.

44.
CALCULUS AND CALCULATORS Example 3

Now that we know what

to look for, we zoom in

(several times) to produce

the first two graphs and

zoom out (several times)

to get the third graph.

Now that we know what

to look for, we zoom in

(several times) to produce

the first two graphs and

zoom out (several times)

to get the third graph.

45.
CALCULUS AND CALCULATORS Example 3

We can read from these

graphs that the absolute

minimum is about -0.02

and occurs when x ≈ -20.

We can read from these

graphs that the absolute

minimum is about -0.02

and occurs when x ≈ -20.

46.
CALCULUS AND CALCULATORS Example 3

There is also a local

maximum ≈ 0.00002 when

x ≈ -0.3 and a local minimum

≈ 211 when x ≈ 2.5

There is also a local

maximum ≈ 0.00002 when

x ≈ -0.3 and a local minimum

≈ 211 when x ≈ 2.5

47.
CALCULUS AND CALCULATORS Example 3

The graphs also show

three inflection points near

-35, -5, and -1 and two

between -1 and 0.

The graphs also show

three inflection points near

-35, -5, and -1 and two

between -1 and 0.

48.
CALCULUS AND CALCULATORS Example 3

To estimate the inflection points closely,

we would need to graph f”.

However, to compute f” by hand is an unreasonable

chore.

If you have a computer algebra system, then it’s easy

to do so.

To estimate the inflection points closely,

we would need to graph f”.

However, to compute f” by hand is an unreasonable

chore.

If you have a computer algebra system, then it’s easy

to do so.

49.
CALCULUS AND CALCULATORS Example 3

We have seen that, for this particular

function, three graphs are necessary to

convey all the useful information.

The only way to display all these features

of the function on a single graph is to draw it

by hand.

We have seen that, for this particular

function, three graphs are necessary to

convey all the useful information.

The only way to display all these features

of the function on a single graph is to draw it

by hand.

50.
CALCULUS AND CALCULATORS Example 3

Despite the exaggerations and distortions,

this figure does manage to summarize

the essential nature of the function.

Despite the exaggerations and distortions,

this figure does manage to summarize

the essential nature of the function.

51.
CALCULUS AND CALCULATORS Example 4

Graph the function f(x) = sin(x + sin 2x).

For 0 ≤ x ≤ π, estimate all (correct to one

decimal place):

Maximum and minimum values

Intervals of increase and decrease

Inflection points

Graph the function f(x) = sin(x + sin 2x).

For 0 ≤ x ≤ π, estimate all (correct to one

decimal place):

Maximum and minimum values

Intervals of increase and decrease

Inflection points

52.
CALCULUS AND CALCULATORS Example 4

First, we note that f is periodic with

period 2π.

Also, f is odd and |f(x)| ≤ 1 for all x.

So, the choice of a viewing rectangle is

not a problem for this function.

First, we note that f is periodic with

period 2π.

Also, f is odd and |f(x)| ≤ 1 for all x.

So, the choice of a viewing rectangle is

not a problem for this function.

53.
CALCULUS AND CALCULATORS Example 4

We start with [0, π] by [-1.1, 1.1].

It appears there are three local maximum values

and two local minimum values in that window.

We start with [0, π] by [-1.1, 1.1].

It appears there are three local maximum values

and two local minimum values in that window.

54.
CALCULUS AND CALCULATORS Example 4

To confirm this and locate them more

accurately, we calculate that

f’(x) = cos(x + sin 2x) . (1 + 2 cos 2x)

To confirm this and locate them more

accurately, we calculate that

f’(x) = cos(x + sin 2x) . (1 + 2 cos 2x)

55.
CALCULUS AND CALCULATORS Example 4

Then, we graph both

f and f’.

Then, we graph both

f and f’.

56.
CALCULUS AND CALCULATORS Example 4

Using zoom-in and the First Derivative Test,

we find the following values to one decimal

Intervals of increase: (0, 0.6), (1.0, 1.6), (2.1, 2.5)

Intervals of decrease: (0.6, 1.0), (1.6, 2.1), (2.5, π)

Local maximum values: f(0.6) ≈ 1, f(1.6) ≈ 1, f(2.5) ≈ 1

Local minimum values: f(1.0) ≈ 0.94, f(2.1) ≈ 0.94

Using zoom-in and the First Derivative Test,

we find the following values to one decimal

Intervals of increase: (0, 0.6), (1.0, 1.6), (2.1, 2.5)

Intervals of decrease: (0.6, 1.0), (1.6, 2.1), (2.5, π)

Local maximum values: f(0.6) ≈ 1, f(1.6) ≈ 1, f(2.5) ≈ 1

Local minimum values: f(1.0) ≈ 0.94, f(2.1) ≈ 0.94

57.
CALCULUS AND CALCULATORS Example 4

The second derivative is:

f”(x) = -(1 + 2 cos 2x)2 sin(x + sin 2x)

- 4 sin 2x cos(x + sin 2x)

The second derivative is:

f”(x) = -(1 + 2 cos 2x)2 sin(x + sin 2x)

- 4 sin 2x cos(x + sin 2x)

58.
CALCULUS AND CALCULATORS Example 4

Graphing both f and f”, we obtain the following

approximate values:

Concave upward on: (0.8, 1.3), (1.8, 2.3)

Concave downward on: (0, 0.8), (1.3, 1.8), (2.3, π)

Inflection points:

(0, 0), (0.8, 0.97),

(1.3, 0.97), (1.8, 0.97),

(2.3, 0.97)

Graphing both f and f”, we obtain the following

approximate values:

Concave upward on: (0.8, 1.3), (1.8, 2.3)

Concave downward on: (0, 0.8), (1.3, 1.8), (2.3, π)

Inflection points:

(0, 0), (0.8, 0.97),

(1.3, 0.97), (1.8, 0.97),

(2.3, 0.97)

59.
CALCULUS AND CALCULATORS Example 4

Having checked that the first figure does

represent f accurately for 0 ≤ x ≤ π, we can

state that the extended graph in the second

figure represents f accurately for -2π ≤ x ≤ 2π.

Having checked that the first figure does

represent f accurately for 0 ≤ x ≤ π, we can

state that the extended graph in the second

figure represents f accurately for -2π ≤ x ≤ 2π.

60.
FAMILIES OF FUNCTIONS

Our final example concerns families

of functions.

As discussed in Section 1.4, this means that

the functions in the family are related to each other

by a formula that contains one or more arbitrary

constants.

Each value of the constant gives rise to a member

of the family.

Our final example concerns families

of functions.

As discussed in Section 1.4, this means that

the functions in the family are related to each other

by a formula that contains one or more arbitrary

constants.

Each value of the constant gives rise to a member

of the family.

61.
FAMILIES OF FUNCTIONS

The idea is to see how the graph of

the function changes as the constant

The idea is to see how the graph of

the function changes as the constant

62.
FAMILIES OF FUNCTIONS Example 5

How does the graph of

the function f(x) = 1/(x2 + 2x + c)

vary as c varies?

How does the graph of

the function f(x) = 1/(x2 + 2x + c)

vary as c varies?

63.
FAMILIES OF FUNCTIONS Example 5

These graphs (the special cases c = 2

and c = - 2) show two very different-looking

These graphs (the special cases c = 2

and c = - 2) show two very different-looking

64.
FAMILIES OF FUNCTIONS Example 5

Before drawing any more graphs,

let’s see what members of this family

have in common.

Before drawing any more graphs,

let’s see what members of this family

have in common.

65.
FAMILIES OF FUNCTIONS Example 5

1

lim 2 =0

x → ±∞ x + 2 x + c

for any value of c, they all have the x-axis

as a horizontal asymptote.

1

lim 2 =0

x → ±∞ x + 2 x + c

for any value of c, they all have the x-axis

as a horizontal asymptote.

66.
FAMILIES OF FUNCTIONS Example 5

A vertical asymptote will occur when

x2 + 2x + c = 0

Solving this quadratic equation, we get:

x = −1 ± 1 −c

A vertical asymptote will occur when

x2 + 2x + c = 0

Solving this quadratic equation, we get:

x = −1 ± 1 −c

67.
FAMILIES OF FUNCTIONS Example 5

When c > 1, there is no vertical

When c > 1, there is no vertical

68.
FAMILIES OF FUNCTIONS Example 5

When c = 1, the graph has a single vertical

asymptote x = -1.

This is because:

1

lim 2

x → −1 x + 2 x + 1

1

= lim

x → −1 ( x + 1) 2

=∞

When c = 1, the graph has a single vertical

asymptote x = -1.

This is because:

1

lim 2

x → −1 x + 2 x + 1

1

= lim

x → −1 ( x + 1) 2

=∞

69.
FAMILIES OF FUNCTIONS Example 5

When c < 1, there are two vertical

asymptotes: x = −1 ± 1 −c

When c < 1, there are two vertical

asymptotes: x = −1 ± 1 −c

70.
FAMILIES OF FUNCTIONS Example 5

Now, we compute the derivative:

2x + 2

f '( x) = − 2 2

( x + 2 x + c)

This shows that:

f’(x) = 0 when x = -1 (if c ≠ 1)

f’(x) > 0 when x < -1

f’(x) < 0 when x > -1

Now, we compute the derivative:

2x + 2

f '( x) = − 2 2

( x + 2 x + c)

This shows that:

f’(x) = 0 when x = -1 (if c ≠ 1)

f’(x) > 0 when x < -1

f’(x) < 0 when x > -1

71.
FAMILIES OF FUNCTIONS Example 5

For c ≥ 1, this means that f increases

on (-∞, -1) and decreases on (-1, ∞).

For c > 1, there is an absolute maximum

value f(-1) = 1/(c – 1).

For c ≥ 1, this means that f increases

on (-∞, -1) and decreases on (-1, ∞).

For c > 1, there is an absolute maximum

value f(-1) = 1/(c – 1).

72.
FAMILIES OF FUNCTIONS Example 5

For c < 1, f(-1) = 1/(c – 1) is a local

maximum value and the intervals of increase

and decrease are interrupted at the vertical

For c < 1, f(-1) = 1/(c – 1) is a local

maximum value and the intervals of increase

and decrease are interrupted at the vertical

73.
FAMILIES OF FUNCTIONS Example 5

Five members of the family are displayed,

all graphed in the viewing rectangle [-5, 4] by

[-2, 2].

Five members of the family are displayed,

all graphed in the viewing rectangle [-5, 4] by

[-2, 2].

74.
FAMILIES OF FUNCTIONS Example 5

As predicted, c = 1 is the value at which

a transition takes place from two vertical

asymptotes to one, and then to none.

As predicted, c = 1 is the value at which

a transition takes place from two vertical

asymptotes to one, and then to none.

75.
FAMILIES OF FUNCTIONS Example 5

As c increases from 1, we see that the

maximum point becomes lower.

This is explained by the fact that 1/(c – 1) → 0

as c → ∞.

As c increases from 1, we see that the

maximum point becomes lower.

This is explained by the fact that 1/(c – 1) → 0

as c → ∞.

76.
FAMILIES OF FUNCTIONS Example 5

As c decreases from 1, the vertical

asymptotes get more widely separated.

This is because the distance between them is 2 1 −c ,

which becomes large as c → – ∞.

As c decreases from 1, the vertical

asymptotes get more widely separated.

This is because the distance between them is 2 1 −c ,

which becomes large as c → – ∞.

77.
FAMILIES OF FUNCTIONS Example 5

Again, the maximum point approaches

the x-axis because 1/(c – 1) → 0 as

c → – ∞.

Again, the maximum point approaches

the x-axis because 1/(c – 1) → 0 as

c → – ∞.

78.
FAMILIES OF FUNCTIONS Example 5

There is clearly no inflection point

when c ≤ 1.

There is clearly no inflection point

when c ≤ 1.

79.
FAMILIES OF FUNCTIONS Example 5

For c > 1, we calculate that

2

2(3 x + 6 x + 4 −c)

f ''( x) = 2 3

( x + 2 x + c)

and deduce that inflection points occur

when x = −1 ± 3(c −1) / 3

For c > 1, we calculate that

2

2(3 x + 6 x + 4 −c)

f ''( x) = 2 3

( x + 2 x + c)

and deduce that inflection points occur

when x = −1 ± 3(c −1) / 3

80.
FAMILIES OF FUNCTIONS Example 5

So, the inflection points become more

spread out as c increases.

This seems plausible from these figures.

So, the inflection points become more

spread out as c increases.

This seems plausible from these figures.