Problem Set 8 Solutions.pdf

8/19/2019 Problem Set 8 Solutions.pdf

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MH1100/MTH112: Calculus I.

Problem Set #9.SOLUTIONS

Problem 1: (#3.1.9 from [Stewart].)

Sketch the graph of a function which is continuous on [1, 5], and has all thefollowing properties:

• It has an absolute maximum at 5.

• It has an absolute minimum at 2.

• It has a local maximum at 3.

• It has local minima at 2 and at 4.

Solutions

There are, of course, very many diff erent ways of doing this problem. Thefollowing is a simple solution:

!1 0 1 2 3 4 5 6 7!2

!1

0

1

2

3

4

5

6

(Note that the problem did not ask us for a diff erentiable solution, in whichcase the above graph would not have been a solution.)

1


2/21


Sketch the graph of a function which is continuous on [1, 5], and has bothof the following properties:

• It has no local maximum or minimum.

• 2 and 4 are critical numbers.

Solution:

Something like the following is what is called for:

!1 0 1 2 3 4 5 6 7!2

!1

0

1

2

3

4

5

6

7

8

This graph shows two diff erent ways that a point can be a critical numberwithout being a local maximum or minimum. (Also note that according toa strict interpretation of the definition of “local max/min”, the boundarypoints here are not local extreme values, because f needs to be defined onboth sides of the point. This isn’t an important issue though.)

2


3/21


(a) Sketch the graph of a function which has two local maxima, one localminimum, and no absolute minimum.

(b) Sketch the graph of a function which has three local minima, two localmaxima, and seven critical numbers.

Solution:

(a) A suitable function is f (x) = x2

x4+1:

!5 0 5!1

!0.5

0

0.5

1

1.5

2

(b) A suitable function is f (x) =

(x + 3π)2 + 1 x ≤ 3π− cos x −3π < x


4/21


Sketch the graph of the function f : [−2, 5) → R given by the rulef (x) = 1 + (x + 1)2,

and use that graph to find and classify the extreme values of f (x).

Solution

This of course is going to be obtained from the standard parabola by ele-mentary geometric operations in the following sequence:

1. Shift the standard parabola one to the left.

2. Shift the result one up.

So we can sketch the function directly:

!6 !4 !2 0 2 4 6!5

0

5

10

15

20

25

30

35

40

45

50

Now we can read off the extrema. They are:

• A local minimum, which is also the absolute minimum of the function,of 1 at x = −1.

Note that because 37, the least upper bound of the function, is not an actualvalue of the function, the function has no absolute maximum. Also, becausex = −2 is a boundary point of the function, it is not a local maximum(strictly speaking).

4


5/21

Problem 5:

Sketch the graph of the function

f (x) =

x2 −1 ≤ x


6/21

Problem 6:

Use the “Closed Interval Method” to find the absolute maximum and theabsolute minimum of the function f (x) = x

x2+1 on the interval [0, 2].

Solution:

The derivative of this function is:

f ′(x) = d

dx

x

x2 + 1

=ddx [x] ∗ (x2 + 1) − x ∗ ddx

x2 + 1

(x2 + 1)2=

x2 + 1 − 2x2(x2 + 1)2

= 1 − x2(x2 + 1)2

.

This function is diff erentiable at every point of this interval, so the criticalnumbers will be the points where this derivative is zero. The derivative willbe zero at x = −1 and x = 1. So the only critical number occurring insidethe given interval is x = 1. So the values we have to compare are:

• 0 , occurring at the left hand end, x = 0.

• 12 , occurring at the critical number x = 1.

• 25 , occurring at the right hand end, x = 2.

Thus the absolute maximum of the function is 12 , occurring at x = 1, andthe absolute minimum of the function is 0, occurring at x = 0.

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7/21

Problem 7: (#3.1.53 from [Stewart])

Find the absolute maximum and the absolute minimum of the functionf (t) = t

√ 4 − t2 on the interval [−1, 2].

Solution:

This function is diff erentiable at every point of its domain except at theright hand end of the interval, t = 2, where it has a vertical tangent. Thederivative is:

f ′(t) = 1 ∗

4 − t2 + t ∗ 12√

4

−t2

∗ ddt

4 − t2

=

4 − t2 − t2√

4 − t2=

1√ 4 − t2

4 − t2 − t2

= 4 − 2t2√

4 − t2 .

Thus the only critical number of the function inside the interval (−1, 2) ist =

√ 2. The values we must compare are:

• −√ 3, occurring at the left hand end, t = −1.• 2, occurring at the critical number t = √ 2.• 0, occurring at the right hand end t = 2.

So we can now read off :

• The absolute minimum of the function is −√ 3, occurring at t = −1.• The absolute maximum of the function is 2, occurring at the critical

number t =√

2.

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8/21


Find the absolute maximum and the absolute minimum of the functionf (t) = 3

√ t(8 − t) on the interval [0, 8].

Solution:

This function is diff erentiable at every point of the interval (0, 8) (though ithas a vertical tangent at the left hand end, 0) so the critical points insidethat interval are the points where the derivative is zero.

The derivative is:

f ′(t) = d

dt 3√ t ∗ (8 − t) +

3√

t∗

d

dt [8

−t]

= 1

3t2

3

∗ (8 − t) + t 13 ∗ (−1)

= 1

t2

3

8

3 − t

3 − t

= t−2

3

8

3 − 4t

3

.

The values we must compare, then, are:

• 0, occurring at the left hand end t = 0.

• 6 3√ 2, occurring at the critical point t = 2.• 0, occurring at the right hand end t = 8.

Thus the absolute maximum is 6 3√

2, occurring at t = 2, and the absoluteminimum is zero, which occurs at both the left hand end as well as the righthand end of the interval.

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Problem 9:

Find the absolute maximum and the absolute minimum of the functionf (x) = x − 2cos x on the interval [−π, π].Solution:

Obviously this function is diff erentiable at every point, so the critical pointswill be the points where the derivative is zero. The derivative is:

f ′(x) = 1 + 2 sin x.

This will be zero when sin x =

−12 . There will be two points x in the interval

[−π, π] where sin x = −12 . They are x = −π6 and x = −5π6 . Thus the valuesof f (x) that we have to compare to find the guaranteed extreme values are:

• f (−π) = 2 − π ≈ −1.142.• f

−5π6 = −5π6 − 2 ∗ cos −5π6 = −5π6 + √ 3 ≈ −0.886.• f

−π6 = −π6 − 2 ∗ cos −π6 = −π6 − √ 3 ≈ −2.25.• f (π) = π − 2cos(π) ≈ 5.142.

Thus the absolute maximum of the function is approximately 5.142, occur-ring at x = π , while the absolute minimum of the function is approximately−2.25, occurring at the critical point x = −π6 .

9


10/21


Show that 5 is a critical number of the function g(x) = 2+ (x − 5)3 but thatg(x) does not have a local extreme value at 5.

Solution:

The derivative of g(x) is

g′(x) = 3(x − 5)2.

Thus 5 is certainly a critical point of the function, because g′(5) = 0. Toshow that x = 5 is not a local extreme value, we can give two explana-

tions. The first is an intuitive explanation, which says: this function is justobtained by translating the standard graph of f (x) = x3 by 5 units to theright and 2 units up. The original function x3 has no extreme values, so thetranslated function 2 + (x − 5)3 will have no extreme values either.

A more precise explanation could use the “Increasing/decreasing” test,which we’ll prove later in this problem set. Note that the derivative

g′(x) = 3(x − 5)2

is positive at all points to the left of x = 5, and is also positive at all pointsto the right of x = 5. So the function is increasing at all points (by the

Increasing/Decreasing test) and so cannot have any extreme values.

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11/21


Recall that a cubic function is a function of the form

f (x) = ax3 + bx2 + cx + d

where a̸ = 0.

(a) Give examples to illustrate that it is possible for a cubic function tohave 2, 1, or no critical numbers.

(b) How many local extreme values is it possible for a cubic function tohave?

Solution:

It suffices to give examples of the three possibilities.

Case I: (No critical numbers). The example f (x) = x3+x is a cubic functionwhose derivative

f ′(x) = 3x2 + 1

is never zero. Because this function is a polynomial, and so diff erentiableat every point, there are no critical numbers. Here is the graph:

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2!8

!6

!4

!2

0

2

4

6

8

Case II: (One critical number). The example f (x) = x3 is a cubicfunction whose derivative f ′(x) = 3x2 is clearly zero at exactly one point,

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x = 0. Here is the graph:

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2!8

!6

!4

!2

0

2

4

6

8

Case III: (Two critical numbers). The example f (x) = x3 − x is a cubicfunction whose derivative is

f ′(x) = 3x2 − 1.This is zero at exactly two point x = 1√

3 and x = − 1√

3. Here is the graph:

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2!8

!6

!4

!2

0

2

4

6

8

What is the situation in general? Well... the derivative of a cubic func-tion is a quadratic function, that is, a function of the form

f ′(x) = Ax2 + Bx + C .

By the “completing the square” procedure, we know that such a function canonly have 0, 1 or 2 roots. So cubics can have at most 2 local extreme values.Part III above shows that there are concrete examples of such functions.

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13/21


Verify that the function f (x) = 3√

x satisfies all the assumptions of the MeanValue Theorem on the closed interval [0, 1]. Then find all the possible pointsc which fit the conclusion of the theorem.

Solution:

This function is certainly continuous on the interval [0, 1]. (Recall thatwe proved explicitly in the -δ part of the course that limx→a x1/3 = a1/3.)

Also, the function is diff erentiable on the open interval (0, 1). (Thisexample has some interest because if the assumptions of the MVT wereslightly diff erent, and required diff erentiability on [0, 1] instead of (0, 1), thenthis function would be disqualified because of the vertical tangent at 0.) Atany rate, this function certainly satisfies the assumptions of the MVT.

So the MVT predicts that there exists a c in (0, 1) where:

f ′(c) = 1

3c−

2

3 = f (1) − f (0)

1 − 0 =3√

1 − 01

= 1.

We can solve this for the only such c, giving:

1

3c−

2

3 = 1

⇒ c2

3 = 1

3

⇒ c2 = 127

⇒ c = 13√

3

⇒ c =√

3

9 .

13


14/21


Consider the function f (x) = 2 − |2x − 1|. Prove that there does not exista point c where

f (3) − f (0) = f ′(c)(3 − 0).Why does this not contradict the Mean Value Theorem?

Solution

To clarify the discussion, we’ll rewrite f (x) explicitly as a piecewise-defined function. When x ≥ 12 , then |2x − 1| = (2x − 1), and when x ≤ 12 ,then |2x − 1| = −(2x − 1). So:

f (x) =

3 − 2x x ≥ 121 + 2x x ≤ 12

.

From this we deduce that the derivative is:

f ′(x) =

−2 x > 12undefined x = 12

2 x < 12

.

Our task is to show that there does not exist a point where f (3)

−f (0) =

f ′(c)(3 − 0). If we rearrange this equation and substitute in values we learnthat such a c would need to satisfy

f ′(c) = f (3) − f (0)

3 − 0 = −3 − 1

3 = −4

3.

Looking at our formula for the derivative we observe that f ′(x) never takesthis value.

This doesn’t contradict the Mean Value Theorem because this functiondoesn’t satisfy all the assumptions of the MVT on the interval [0 , 3]. It isdefinitely continuous on [0, 3], so it satisfies the first assumption, but it is

not diff erentiable on (0, 3), because it is not diff erentiable at x = 12 . Thus it

fails the second assumption.

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15/21

Problem 14:

Apply standard theorems of calculus to deduce that equation

1 + 2x + x3 + 4x5 = 0

has exactly one real root.

Solution:

To begin, we observe that the function f (x) = 1 + 2x + x3 + 4x5 has atleast one real root. (Recall that a “real root” means a real number xwhere f (x) = 0. We say “real” root because functions can have more roots

over the complex numbers.) The reason it has at least one root is basicallybecause the leading term is an odd power of x, x5. Alternatively, we canuse the IVT, noting that:

• f (0) = 1 > 0

• f (−1) = −6 < 0• f (x) is continuous on [−1, 0].• So the IVT says that f has a root in (−1, 0).That was the easy bit. Next we need to show that there cannot be a

second root. How to show this? Well, assume that there were two roots, x1and x2. It would then follow from Rolle’s theorem (because this polynomial,of course, is diff erentiable everywhere) that there exists a point c betweenx1 and x2 where f

′(c) = 0. But the derivative is:

f ′(x) = 20x4 + 3x2 + 2

and will never be zero. So the second root is impossible.

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16/21


Apply standard theorems of calculus to deduce that the equation

x4 + 4x + c = 0

has at most two real roots (for any choice of constant c).

Solution:

To show that this equation has at most two real roots, we’ll assumethat there are three real roots, x1 < x2 < x3, (i.e. f (x1) = 0, etc.) and we’lldeduce an impossibility from that, using Rolle’s theorem.

Well, assume that there were three such roots. Looking at the pair x1and x2, Rolle’s theorem says that there exists a point, let’s call it x12 suchthat x1 < x12 < x2 and such that

f ′(x12) = 0.

Furthermore, applying Rolle’s theorem to the pair x2 and x3, we learn thatthere will exist a point x23 such that x2 < x23 < x3 where

f ′(x23) = 0.

Lets summarize our deductions so far: if the original function has three

roots, then Rolle’s theorem says that the derivative

f ′(x) = 4x3 + 4

has at least two roots.But that is impossible! The derivative of the derivative,

(f ′)′(x) = 12x2

is only non-positive at a single point, so it must be an increasing function.It is impossible for an increasing function to have two roots.

So our original assumption, that f

(x

) has three

roots, was impossible.(An alternative way to complete the deduction here is to observe that

the function x3 is increasing, which means that the function 4x3 + 4 will alsobe increasing, so it must be 1-1.)

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17/21


Assume that some function satisfies 3 ≤ f ′(x) ≤ 5 for all values of x. Showthat

18 ≤ f (8) − f (2) ≤ 30.Solution:

This is just the Mean Value Theorem applied for this f (x) to the interval[2, 8]. The MVT says that there will exist a c ∈ (2, 8) such that

f ′(c) = f (8) − f (2)

8

−2

.

In other words, for some c ∈ (2, 8):

f (8) − f (2) = 6f ′(c).

Now we were told that for every possible x, the derivative f ′(x) lay be-tween 3 and 5. So:

3 ≤ f ′(c) ≤ 5⇒ 18 ≤ 6f ′(c) ≤ 30⇒ 18 ≤ f (8) − f (2) ≤ 30.


Does there exist a function f (x) such that f (0) = −1, f (2) = 4, and f ′(x) ≤2 for all x?

Solution:

No, there cannot! Why? According to the MVT there would have to exista c ∈ (0, 2) such that

f ′(c) = f (2)− f (0)

2 − 0 = 52

> 2.

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18/21


Suppose that f is an odd function which is diff erentiable at every point.Prove that for every positive number b, there exists a number c ∈ (−b, b)such that f ′(c) = f (b)b .

Solution:

We apply the MVT to the interval [−b, b], and deduce that there existsa c ∈ (−b, b) where

f ′(c) = f (b) − f (−b)

b − (−b) .

The function is odd, so this c has the property that:

f ′(c) = f (b) − f (−b)

2b =

f (b) + f (b)

2b =

f (b)

b .


Use the Mean Value Theorem to prove the inequality

| sin x − sin y| ≤ |x − y|for all x and y .

Solution:

Assume that a < b. (Note that

“| sin a − sin b| < |a − b|” if and only if “| sin b − sin a| < |b − a|”so this case implies the case b < a.)

Applying the MVT we learn that there exists a c such that:

sin a − sin ba − b = cos(c).

This implies that:

sin a − sin ba − b = |cos c| ≤ 1.

Thus:|sin a − sin b| ≤ |a − b|

as was required.

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19/21

Problem 21: (Based on #3.3.30 from [Stewart])

Use the Mean Value Theorem to prove that if a function satisfies f ′(x) = cfor all x, where c is some constant, then there exists a constant d such that

f (x) = cx + d.

Solution

Define a functiong(x) = f (x) − cx.

By the rules for diff erentiation, this function is diff erentiable at every pointwith derivative:

g′(x) = f ′(x) − c = c − c = 0.In the lectures we used the MVT to prove that any function whose

derivative is zero at every point is a constant function. Thus there existssome constant d such that

g(x) = d.

Substituting g(x) = f (x) − cx concludes the proof.


A fixed point of a function f (x) is a number a such that f (a) = a. Let f (x)be a function which is diff erentiable at every point, and which satisfies theproperty that f ′(x) is never 1. Prove that such a function cannot have morethan one fixed point.

Solution

Assume that this function f (x) has two diff erent fixed points: an a1where f (a1) = a1, and an a2 where f (a2) = a2. We’ll show that this impliesthat f (x) must have its derivative equal to 1 somewhere, which would be acontradiction.

To do this we basically just apply the MVT to the interval [ a1, a2]. Ittells us that there exists a point c where

f ′(c) = f (a2) − f (a1)

a2 − a1=

a2 − a1a2 − a1

= 1.

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20/21

Problem 23: (The increasing/decreasing test.)

Let f (x) be a function which is continuous on some closed interval [a, b],and diff erentiable on (a, b). Prove that:

(a) If f ′(x) > 0 for all x ∈ (a, b), then f (x) is increasing on [a, b].(b) If f ′(x) < 0 for all x ∈ (a, b), then f (x) is decreasing on [a, b].

Solution

We’ll focus on the proof of the “increasing” case: Part (a). The decreas-ing case is exactly the same, but with a few signs changed.

Let x1 < x2 be an arbitrary pair of points from [a, b]. Our task is toshow that f (x1) < f (x2).

Note that the function f (x) satisfies the assumptions of the MVT on[x1, x2], which are that

• f (x) is continuous on [x1, x2].

• f (x) is diff erentiable on (x1, x2).

The Mean Value Theorem concludes that there exists a point c ∈ (x1, x2)where

f ′(c) = f (x2) − f (x1)x2 − x1 .

Rearranging this we get:

f (x2) − f (x1) = f ′(c) ∗ (x2 − x1).

This is greater than zero because x1 < x2, and because f ′(c) > 0 (because

c must lie in (a, b)).

Thus f (x1) < f (x2).

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Problem 24⋆: (#11.26 from [Spivak])

Let f be a function and M > 0 a real number such that f ′(x) ≥ M > 0for all x ∈ [0, 1]. Prove that there is an interval of length 14 in [0, 1] on which|f (x)| ≥ M 4 .

Solution

Our task is to prove the existence of an interval of length 14 in [0, 1] whereeither f (x) ≤ −M 4 or f (x) ≥ M 4 . (It may help to draw a picture here tovisualize this).

The first thing to observe is that because the derivative is always > 0,

this f (x) is an increasing function (by the Increasing Test proved in theprevious problem).

One approach is to split the deduction into 2 cases, depending on thevalue the function takes at the point 1/4.

Case 1: f 14

≤ −M 4 .Because f (x) is an increasing function, this implies that f (x) ≤ −M 4 for

every x ∈ 0, 14. So the interval x ∈ 0, 14 is the required interval.Case 2: f 14 > −

M 4 .

The way we’ll deal with this case is to use the MVT to deduce thatf 34

≥ M 4 . Once we know that then, because we know that f (x) is increas-ing, we can conclude that the interval

34 , 1

has the required properties.

To deduce f 34

≥ M 4 :By the MVT applied to the interval

14 ,

34

, we know that there exists a

point c ∈ 14 , 34 such thatf

3

4

− f

1

4

= f ′(c) ∗

3

4 − 1

4

≥ M ∗ 1

2 .

So because in this case f 14

> −M 4 we deduce that

f

3

4

≥ f

1

4

+

M

2 > −M

4 +

M

2 =

M

4 .

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