problem set 6 solutions.pdf

8/19/2019 Problem Set 6 Solutions.pdf

http://slidepdf.com/reader/full/problem-set-6-solutionspdf 1/21

MH1100/MTH112: Calculus I.

Problem list for Week #7.Solutions.

Problem 1:

Use the definition of the derivative to determine the derivative function of

f (x) = x3 − 3x + 5.

State the domain of the function and the domain of the derivative.

Solution

The derivative of f (x) at a point x is given by the following limit, if it exists:

f (x) = limh→0

f (x + h) − f (x)

h

= limh→0

(x + h)3 − 3(x + h) + 5

− (x3 − 3x + 5)

h

= limh→0

x3 + 3x2h + 3xh2 + h3 − 3x − 3h + 5 − x3 + 3x − 5

h

= limh→0

3x2h + 3xh2 + h3 − 3h

h

= limh→0

3x2 + 3xh + h2 − 3

= 3x2 − 3.

The function f (x) is a polynomial, so its domain is all of R. And we justsaw that the limit which defines the derivative exists at every point x ∈ R,so its domain is all of R as well.

1



Problem 2 (#2.2.25 from [Stewart]):


g(x) =√

9 − x.


Solution

The domain of f is clearly (−∞, 9].

The domain of the derivative will be some subset of (−∞, 9). (f can’t bedifferentiable at a = 9 because f needs to be defined on both sides of a pointa to have a derivative at that point.)

So let x ∈ (−∞, 9). According to the definition

f (x) = limh→0

f (x + h) − f (x)

h

= limh→0

9 − (x + h) −√

9 − x

h

= limh→0

9 − (x + h) −√

9 − x

h ∗

9 − (x + h) +

√ 9 − x

9 − (x + h) +√

9 − x

= limh→0

9 − (x + h)2 − √ 9 − x2h

9 − (x + h) +√

9 − x

= limh→0

−h

h

9 − (x + h) +√

9 − x

= − 1

2√

9 − x.

This limit exists for every x < 9, so the domain of f (x) is all of (−∞, 9).

2





f (x) = x2 − 1

2x− 3.


Solution

The domain of f (x) is R\32

. The domain of the derivative will be this set

or some subset of it.

Let x

∈R

\3

2. Then:

f (x)

= limh→0

f (x + h) − f (x)

h

= limh→0

(x+h)2−12(x+h)−3 − x2−1

2x−3

h

= limh→0

(x2 + 2hx + h2) − 1

(2x − 3) − (x2 − 1)(2x + 2h − 3)

h(2(x + h) − 3)(2x − 3)

= limh→0

(x2 − 1)(2x − 3) + (2hx + h2)(2x − 3) − (x2 − 1)(2x − 3) − 2h(x2 − 1)

h(2(x + h)

−3)(2x

−3)

= limh→0

2x(2x − 3) + h(2x − 3) − 2(x2 − 1)(2(x + h) − 3)(2x − 3)

= 2x(2x − 3) − 2(x2 − 1)

(2x − 3)2

= 2x2 − 6x + 2

(2x − 3)2 .

This limit exists for every x ∈ R\32

, so that is the domain of the derivative.

3




Given a certain function f (x), we have plotted f (x), f (x), and f (x). Matchthese functions to the different graphs a, b and c. Justify your answer.

Solution

I think the way to do this is to work out which of the displayed graphs canbe the derivative of which. The derivative of Graph a can only be Graphb, because Graph c is already negative to the left of the y-axis, but Grapha is not decreasing there. Similarly the derivative of Graph b can only beGraph c (for example, the maximum of b occurs at exactly the point wherec goes from positive to negative).

Thus: Graph a depicts f , Graph b depicts f , and Graph c depicts f .


Given a certain function f (x), we have plotted f (x), f (x), f (x), and f (x).Match these functions to the different graphs a, b, c and d. Justify youranswer.

Solution

Note that the derivative of the function drawn in Graph d seems to be givenby Graph c. (For example, you can match the turning points of d to thepoints where the Graph c passes from positive to negative, and vice versa).The derivative of c could be given by b (note that c decreases at exactly thepoints where b is negative, and vice versa). And finally the derivative of bcould be given by a.

So the following seems to be a consistent match-up: Graph d depicts f ,Graph c depicts f , Graph b depicts f , and Graph a depicts f .

4




Consider the function g(x) = x2/3.

(i) Show that g(x) is not differentiable at x = 0.

(ii) Calculate g(a) in the case that a = 0.

(iii) Show that the curve y = x2/3 has a vertical tangent line at (0, 0).

(iv) Illustrate by graphing g(x) and g(x).

Solution to (i).

To show that g(x) is not differentiable at x = 0 we have to show that thelimit which defines the derivative at that point, namely

limh→0

g(0 + h) − g(0)

h ,

does not exist.The calculation is:

limh→0

g(0 + h) − g(0)

h = lim

h→0

h2/3 − 0

h = lim

h→0h−1/3

which obviously does not exist. (It approaches +

∞ as h approaches 0 from

the right and −∞ as h approaches 0 from the left.)

5



Solution to (ii).

Let a = 0. Then:

g(a) = limh→0

g(a + h) − g(a)

h

= limh→0

(a + h)2/3 − a2/3

h

= limh→0

(a + h)2/3 − a2/3

h ∗ (a + h)4/3 + (a + h)2/3a2/3 + a4/3

(a + h)4/3 + (a + h)2/3a2/3 + a4/3

= limh→0

(a + h)2/3

3 − a2/3

3h (a + h)4/3 + (a + h)2/3a2/3 + a4/3

= limh→0

(a + h)2 − a2

h

(a + h)4/3 + (a + h)2/3a2/3 + a4/3

= limh→0

(a + h − a)(a + h + a)

h

(a + h)4/3 + (a + h)2/3a2/3 + a4/3

= limh→0

(2a + h)

(a + h)4/3 + (a + h)2/3a2/3 + a4/3

= 2a

3a4/3

= 2

3a−1/3.

Solution to (iii).

According to the textbook, to check that a graph has a vertical tangent lineat a point, we have to check that it is

(a) continuous at that point, and that

(b) limx→a |g(x)| = ∞.

The given g(x) satisfies:

limx→0

g(x) = limx→0

x2/3 = 0 = g(0)

so it is continuous at the point 0. Also, using the answer to (ii) we calculate

limx→0

|g(x)| = limx→0

2

3x−1/3

= +∞.

So this graph has a vertical tangent at x = 0.

6



Solution to (iv).

The graphs of g(x) and g(x) are:

7




Using the definition of derivative, determine the derivative function of thefunction

f (x) = |x− 6|.Sketch the graph of f (x) and of f (x).

Solution

This function is defined by different rules, depending on whether x < 6,x = 6 or x > 6. The rules are:

f (x) = −(x− 6) x < 6,

0 x = 6,

x − 6 x > 6.

To calculate the derivative function, we are going to have to split this into3 cases.

Case 1: a > 6.

In this case, f (x) is given by the rule x − 6 at every point close to a, so thecalculation becomes:

limh→0

f (a + h) − f (a)

h

= limh→0

(a + h − 6) − (a − 6)

h

= limh→0

h

h

= 1.

Case 2: a < 6.

In this case, f (x) is given by the rule −(x− 6) at every point close to a, sothe calculation becomes:

limh→0

f (a + h) − f (a)

h = lim

h→0

−(a + h − 6) − (−(a − 6))

h

= limh→0 −h

h

= −1.

8



Case 3: a = 6. The limit which determines the derivative at a = 6 is:

limh→0

f (6 + h) − f (6)

h .

The function inside the limit is defined by two different rules, depending onwhether h > 0 or h < 0, so to understand this limit we’ll need to considerthe corresponding 1-sided limits. The limit from the right is:

limh→0+

f (6 + h) − f (6)

h = lim

h→0+

((6 + h) − 6) − 0

h = lim

h→0+

h

h = 1.

The limit from the left is:

limh→0−

f (6 + h) − f (6)h

= limh→0−

(−(6 + h) + 6) − 0h

= limh→0+

−hh

= −1.

The 1-sided limits are different, so the derivative is not defined at the pointx = 6.

In summary:

f (x) =

1 x > 6undefined x = 6

−1 x < 6

The graphs of f (x) and f (x) are:

9



Problem 8:

Using the definition of derivative, determine the derivative function of thefunction

f (x) = |x2 − 9|.Sketch the graph of f (x) and of f (x).

Solution

To begin, note that x2 − 9 = (x − 3)(x + 3), so:

(x2 − 9)

> 0 if x < −3,0 if x =

−3,

< 0 if −3 < x < 3,0 if x = 3,

> 0 if x > 3.

Thus:

f (x) =

x2 − 9 if x < −3,0 if x = −3,

−(x2 − 9) if −3 < x < 3,0 if x = 3,

x2 − 9 if x > 3.

To determine the derivative function, we should consider these 6 cases sep-

arately.Case 1: x < −3.

In this case the calculation is:

f (x) = limh→0

f (x + h) − f (x)

h

= limh→0

((x + h)2 − 9) − (x2 − 9)

h

= limh→0

2hx + h2

h= 2x.

Case 2: x = −3.

In this case the limit we need to understand is

limh→0

f (−3 + h) − f (−3)

h .

10



The function f is defined by different rules on the two sides of −3, so the

function f (−3+h)−f (−3)

h is defined by two different rules on the two sides of h = 0. So to understand this limit, we should investigate the 1-sided limitsat h = 0. The limit from the left is:

limh→0−

f (−3 + h) − f (−3)

h = lim

h→0−

(−3 + h)2 − 9 − (0)

h = −6.

The limit from the right is:

limh→0+

f (−3 + h) − f (−3)

h = lim

h→0+

−((−3 + h)2 − 9) − (0)

h = 6.

Thus, because the 1-sided limits are different, the limit limh→0f (−3+h)−f (−3)

hdoes not exist, and so the function f is not differentiable at x = −3.

The other cases are all small modifications of these 2 cases. The finalanswer is:

f (x) =

2x if x < −3,undefined if x = −3,

−2x if −3 < x < 3,undefined if x = 3,

2x if x > 3.

The graphs of f (x) and f (x) are:

11




Consider the function f (x) = x|x|.(i) Graph this function.

(ii) For what values of x is f (x) differentiable?

(iii) Find a formula for f (x).

Solution

This function is determined by different rules in different regions accordingto:

f (x) = −x2 if x < 0,

0 if x = 0,x2 if x > 0.

The graph of f is straightforward to assemble using the standard quadraticgraph. It is:

This graph appears to have a non-vertical tangent line at every point, sowe expect that the function will be differentiable at every point. To answerwith certainty, we’ll apply the definition of derivative to calculate it.

12



There are three cases to consider.

Case 1: x < 0. In this case the calculation is:

f (x) = limh→0

f (x + h) − f (x)

h = lim

h→0

−(x + h)2 − (−x2)

h = −2x.

Case 2: x > 0. In this case the calculation is:

f (x) = limh→0

f (x + h) − f (x)

h = lim

h→0

(x + h)2 − (x2)

h = 2x.

Case 3: x = 0. In this case, to determine the derivative at x = 0 the limitwe have to understand is:

f (0) = limh→0

f (0 + h) − f (0)h

.

Because f (x) is defined by two different rules on the two sides of x = 0,

the expression f (0+h)−f (0)h is defined by 2 different rules on the two sides of

h = 0. This means that we need to consider the 1-sided limits at h = 0.The limit from the left is:

limh→0−

f (0 + h) − f (0)

h = lim

h→0−

h2 − 0

h = 0.

And the limit from the right is:

limh→0+

f (0 + h) − f (0)h

= limh→0+

−h2 − 0h

= 0.

The two 1-sided limits both exist, and they are equal, which means thatthe limit limh→0

f (0+h)−f (0)h exists and equals 0. Thus the function f (x) is

indeed differentiable at x = 0, with derivative 0.In summary, the function f (x) is differentiable at every point, with

derivative given by:

f (x) =

−2x if x < 0,0 if x = 0,

2x if x > 0.

(Note that a short way of writing this result is: f (x) = 2|x|.)

13



Problem 10 (The rational case of the power rule.):

(i) Let q ∈ N, and consider f (x) = x1/q. Assume that a > 0. Use thedefinition of derivative to determine f (a).

(ii) Let p, q ∈ N, and consider f (x) = x p/q . Assume that a > 0. Use theresult of part (i) and the product rule to determine f (a).

Solution to (i).

The trick here is to use an appropriate version of the “really usefulidentity” to get rid of the q -th roots on the top line. The version we will usebelow is:

(t1/q

− s1/q

)t1/qq−1 + t

1/qq−2 s1/q1 + . . . + s

1/qq−1 = t − s.

When we use this below, we will use summation notation to make thisexpression more compact:

(t1/q − s1/q)

q−1i=0

t1/q

i s1/q

q−1−i= t − s.

The computation is:

f (a) = limh→0

(a + h)1/q − a1/q

h

= limh→0

(a + h)1/q

− a1/q

h ∗ q−1

i=0 (a + h)1/qi a

1/qq−1−iq−1i=0

(a + h)1/q

i a1/q

q−1−i

= limh→0

(a + h)1/q

q − a1/q

qhq−1

i=0

(a + h)1/q

i a1/q

q−1−i

= limh→0

(a + h) − a

hq−1

i=0

(a + h)1/q

i a1/q

q−1−i

= 1

limh→0

q−1i=0

(a + h)1/q

i a1/q

q−1−i

=

1q−1i=0

a1/qi a1/qq−1−i

= 1

q a1q−1

.

This result is what is required for the power rule to be true for the case x1/q.

14



Solution to (ii).

In this part we have to use the result of (i), and the product rule, to deter-mine the derivative of f (x) = x p/q , where p and q are positive integers.

To do this, we just rewrite f (x) in the following way:

f (x) = x p/q = x1/q . . . x1/q p

.

According to part (i), the factor x1/q is differentiable at a, so by theproduct rule we deduce that f (x) is also differentiable at a, with derivative

f

(a) = ddx x

1/q

. . . x1/q

p

= p d

dx

x1/q

x1/q

p−1(By the product rule.)

= pq x

1q−1

xp−1

q (By part (i).)

= pq x

p

q−1

This confirms the power rule for this case.

15




The left-hand derivative and the right-hand derivative of a functionf (x) at a point a are defined by

f −

(a) = limh→0−

f (a + h) − f (a)

h and f +(a) = lim

h→0+

f (a + h) − f (a)

h

if these limits exist.

(i) Briefly explain why f is differentiable at a if and only if both of these1-sided derivatives exist and are equal.

(ii) Now consider the function

f (x) =

0 if x ≤ 0,5 − x if 0 < x < 4,1

5−x if x ≥ 4.

Determine f −

(4) and f +(4).

(iii) Sketch the graph of f (x).

(iv) Where is f discontinuous?

(v) Where does f fail to be differentiable?

Solution to (i).

According to the definition, a function f is differentiable at a point a if and only if the following limit exists

limh→0

f (a + h) − f (a)

h .

And according to a standard theorem about limits, this limit exists if andonly if the two corresponding 1-sided limits

limh→0−

f (a + h) − f (a)

h

and limh→0−

f (a + h) − f (a)

h

exist and are equal. These 2 limits are the “1-sided derivatives” introducedabove.

16



Solution to (ii).

The left-hand derivative of f (x) at the point 4 is given by the limit:

f −

(4) = limh→0−

f (4 + h) − f (4)

h

= limh→0−

(5 − (4 + h)) − 15−4

h = lim

h→0−

−h

h = −1.

And the right-hand derivative of f (x) at the point 4 is given by the limit:

f +(4) = limh→0+

f (4 + h) − f (4)

h

= limh→0+

1

5−

(4+h) − 1

5−

4h = lim

h→0+h

h(1 − h) = 1.

Solution to (iii), (iv), and (v).

Here is the graph:

The function is discontinuous at 0 (where it has a jump discontinuity)and at 5 (where it has an infinite discontinuity).

The function fails to be differentiable at 0 and 5 (because it fails to even

be continuous there) and also fails to be differentiable at x = 4, where it hasa corner. (To be precise, at x = 4 the left and right derivatives exist but areunequal.)

17




Use the definition of derivative to prove that:

(i) The derivative of an even function is an odd function.

(ii) The derivative of an odd function is an even function.

Also, try and understand these facts using your geometric intuition and byconsidering a few concrete examples.

Solution to (i).

In this solution we’ll use the fact that for any function F (x):

limh→0

F (h) = limh→0

F (−h) ().

This statement should be clear to the intuition (draw a picture), and is alsofairly straightforward to prove using the -δ definition of limit.

Let f (x) be an even function, and assume that it has a derivative at somepoint a. We’ll show that it also has a derivative at −a, and f (−a) = −f (a).

f (−a) = limh→0f (−a+h)−f (−a)

h (By defn. of derivative.)

= limh→0f (a−h)−f (a)

h (Because f is even.)

= − limh→0f (a−h)−f (a)

−h

= − limh→0f (a+h)−f (a)

h (Using ().)

= −f (a). (Because f is differentiable at a.)

18



Solution to (ii).

Let f (x) be an odd function, and assume that it has a derivative at somepoint a. We’ll show that it also has a derivative at −a, and also thatf (−a) = f (a).

f (−a) = limh→0f (−a+h)−f (−a)

h

= limh→0−f (a−h)+f (a)

h (Because f is odd.)

= − limh→0f (a−h)−f (a)

h

= + limh→0f (a−h)−f (a)

−h

= + limh→0f (a+h)−f (a)

h (Using ().)

= +f (a). (Because f is differentiable at a.)

19



Problem 13 (related to #5.9 from [Spivak]):

Let F (x) be a function and let a ∈ R. Use the -δ definition of limit tocarefully prove that

limx→a

F (x) = limh→0

F (a + h).

Solution

For precision we’ll split this into two parts.

Part A: In this part we’ll assume that the limit limx→a F (x) exists, andequals some number K , and we’ll use that information to show that theother limit limh→0 F (a + h) exists as well, and is also equal to K .

The assumption in Part A is: there exists a rule δ 1() satisfying theproperty that for every > 0, whenever x satisfies 0 < |x − a| < δ 1() then|F (x) − K | < ().

Using this information we’ll now prove that limh→0 F (a + h) = K .

Proof:

Let be an arbitrary positive real.

Set δ = δ 1().

Let h be an arbitrary real such that 0 < |h − 0| < δ .

For this h, |(a + h) − a| = |h| < δ .

It follows from () that for this h, |F (a + h) − K | < .

Part B : This part will be the same thing but going backwards. We’llassume that limh→0 F (a + h) exists, and equals some L, and then we’ll usethat information to prove that limx→a F (x) also exists, and also equals L.

The assumption in Part B is: there exists a rule δ 2(), such that wheneverh satisfies 0 < |h| < δ 2(), then |F (a + h) − L| < ().

Now we’ll prove limx→a F (x) = L.

Proof:

Let be an arbitrary positive real.

Set δ = δ 2().

Let x be an arbitrary real such that 0 <

|x

−a

| < δ .

By (), for this x, |F (x) − L| = |F (a + (x − a)) − L| < .

20



Problem 14 (#9.14 from [Spivak]):

Consider the function

f (x) =

x2 if x is rational,0 if x is irrational.

Prove that f (x) is differentiable at the point x = 0.

Solution

Our task is to show that the following limit exists:

limh→0

f (0 + h) − f (0)

h

.

Let F (h) be the function inside this limit: F (h) = f (0+h)−f (0)

h . We needto show that limh→0 F (h) exists. Note that F (h) is given by the formula:

F (h) =

h if h is rational,0 if h is irrational.

It is clear that limh→0 F (h) exists and equals 0 because we can squeezeit between −|h| and +|h|. Thus: f (0) = 0.

21

problem set 6 solutions.pdf

Documents