MATH 100, Section 110 (CSP)
Weeks 9 and 10: Marked Homework Solutions
2010 Nov 18
1. [10] Let f (x) = 1 + 2x + x3 + 4x5 . We are supposed to prove that f (x) = 0 has exactly
one solution in all of −∞ < x < ∞. I.e. there is exactly one number c in (−∞, ∞) such
that f (c) = 0.
Observe that f is a polynomial, so it is continuous and differentiable everywhere.
We first prove there is at least one solution. Substituting in x = 0, it is easy to see f (0) = 1,
which is positive. Since the term with the largest power is 4x5 , for negative enough values
of x, f (x) will have the same sign as 4x5 , which will be negative. So we try some negative
values of x. Trying x = −1, for example, gives f (−1) = −6, which is negative. Since
f is continuous on the closed interval [−1, 0], the hypotheses of the Intermediate Value
Theorem have been verified and we can conclude that there is at least one value c, between
−1 and 0, such that f (c) = 0.
Next, we prove that there can be no more than one solution, using proof by contradiction
(or reductio ad absurdum). Suppose there are two (or more) solutions of f (x) = 0. Then we
can choose two solutions, and one must be larger than the other. Label these solutions as a
and b, with a < b. Because they are solutions of f (x) = 0, we know these specific numbers
satisfy f (a) = 0 and f (b) = 0. Since f is continuous everywhere, it is continuous on [a, b].
And since f is differentiable everywhere, it is differentiable on (a, b). The hypotheses of the
Mean Value Theorem have been verified, so now we can conclude that there is a number
c̃ (strictly between a and b, but its location doesn’t matter to us) such that
f 0 (c̃)(b − a) = f (b) − f (a) = 0 − 0
and since (b − a) is nonzero (why?), we must have
f 0 (c̃) = 0.
On the other hand, taking the derivative of f explicitly, we get
f 0 (x) = 2 + 3x2 + 20x4
which is always ≥ 2 due to the even powers of x. So f 0 (x) = 0 is impossible. This is a
contradiction, so the underlined statement above must be false. Therefore there can be no
more than one solution of f (x) = 0.
In summary, we have proved there is at least one solution, and there can be no more than
one solution. Therefore there is exactly one solution.
2. [10] (a) Since f is continuous on [1, 2] and is differentiable on (1, 2) the Mean Value
(1)
Theorem implies that there is some c, with 1 < c < 2, such that f 0 (c) = f (2)−f
> 0.
2−1
(b) By the Mean Value Theorem, there is some c in (1, 2) such that f (2)−f (1) = f 0 (c)(2−1)
so f (2) = 5 + f 0 (c). By hypothesis, f 0 (c) is between 4 and −4, so f (2) is between 5 + 4 = 9
and 5 − 4 = 1.
1
It is possible for f (2) to equal 1 or 9. For example, consider the (different) functions f (x) =
4(x − 1) + 5 or f (x) = −4(x − 1) + 5. Both of these examples are continuous on [1, 2],
are differentiable on (1, 2), and satisfy the given conditions f (1) = 5, |f 0 (x)| ≤ 4 for all
1 < x < 2. In the first example f (2) = 9 and in the second example f (2) = 1.
3. [8] Note that the graph of a function f that satisfies the three properties i)–iii) is simply
the graph of a smooth function that joins the points (0, 0) and (1, 1).
For any function f that satisfies the three properties i)–iii), let
M (f ) = the maximum of |f 0 (x)| on (0, 1).
By the Mean Value Theorem, for any function f satisfying the three properties i)–iii), there
(0)
is some c in (0, 1) such that f 0 (c) = f (1)−f
= 1 so the maximum of |f 0 (x)| on (0, 1) has to
1−0
be at least that value of 1:
M (f ) ≥ 1 for any function f satisfying properties i)–iii).
The specific function f0 (x) = x satisfies properties i)–iii) and the maximum of |f00 (x)| on
(0, 1) is exactly equal to 1:
M (f0 ) = 1 for a specific function f0 satisfying properties i)–iii).
From the above two facts, we could guess that 1 is the smallest possible value of M (f ). (You
can also draw a few pictures of graphs of functions that satisfy properties i)–iii). Any graph
that is not a straight line joining (0, 0) and (1, 1) always seems to have some points where
the absolute value of the slope of the tangent line is strictly greater than 1.)
We have to prove that the guess is correct. Let f be any function satisfying properties i)–iii)
that is not equal to f0 (i.e. its graph is not a straight line joining (0, 0) and (1, 1).) Then
f (x) 6= x for some x in (0, 1).
Case 1: If f (x) > x, then apply the Mean Value Theorem on the interval [0, x] to
(0)
=
conclude that there is some c in (0, x) (and c is therefore in (0, 1)) with f 0 (c) = f (x)−f
x−0
f (x)
> 1.
x
Case 2: If f (x) < x, observe that 1 − f (x) > 1 − x and apply the Mean Value Theorem
on the interval [x, 1] to conclude that there is some c in (x, 1) (and c is therefore in (0, 1))
(x)
(x)
= 1−f
> 1.
with f 0 (c) = f (1)−f
1−x
1−x
See the figure below (the dashed lines are y = x, x in [0, 1]).
2
Cases 1 and 2 together imply is that if f is any function satisfying i)–iii) and f 6= f0 , then
there is some number c in (0, 1) so that |f 0 (c)| is strictly greater than 1, and therefore the
maximum of |f 0 (x)| for all numbers x in (0, 1) must be strictly greater than 1. So we have
M (f ) > 1 if f 6= f0 , and M (f0 ) = 1. Therefore the smallest possible value of M (f ) is 1,
obtained precisely when the function is f (x) = f0 (x) = x.
(This question came from an old MATH 100 final exam.)
4. (a) [3] An equation for L is
y = f (a) + f 0 (a)(x − a),
which for f (x) = 1 + ex becomes
y = 1 + ea + ea (x − a).
Since x = 0, y = 0 is known to be on this line, we have
0 = 1 + ea + ea (0 − a),
or
(b) [5] Let
0 = 1 + ea − aea .
g(x) = 1 + ex − xex ,
and we have to prove that g(x) = 0 has exactly one solution (which we called a above).
Observe that g is continuous and differentiable everywhere.
Trying some different x-values (a sketch of the tangent line situation described above would
help a lot), we get
g(0) = 2
which is positive, and
g(2) = 1 − e2
which is negative (recalling that e > 1). Since g is continuous everywhere, it is continuous
on [0, 2], and the Intermediate Value Theorem can be used. We conclude that there is
at least one value of a (between 0 and 2) such that g(a) = 0.
Let us compute
Notice that
g 0 (x) = −xex .
g 0 (x) < 0 for all 0 < x < ∞,
so we can use the familiar Mean Value Theorem argument (question 2) to prove that
there can be no more than one solution in (0, ∞): suppose there were two solutions, 0 <
a1 < a2 < ∞, then g is continuous on [a1 , a2 ] and differentiable on (a1 , a2 ), and there exists
c between a1 and a2 such that g 0 (c) = (g(a2 ) − g(a1 ))/(a2 − a1 ) = (0 − 0)/(a2 − a1 ) = 0. But
such a c would have to satisfy 0 < c < ∞ since it is between the two positive numbers a1
and a2 , and this contradicts the fact that g 0 (x) < 0 for all 0 < x < ∞.
3
But what about (−∞, 0], could there be any roots in there? To answer this, we look at g(x)
directly, not g 0 (x).
g(x) = 1 + ex − xex ≥ 1 + ex ≥ 1
for any −∞ < x ≤ 0, because in that case −xex ≥ 0 (ex is always positive). So g(x) = 0 is
impossible if −∞ < x ≤ 0.
Summarizing, g(x) = 0 has at least one solution, can have no more than one solution in
(0, ∞) and can have no solution in (−∞, 0]. Therefore g(x) = 0 has exactly one solution in
the real line (−∞, ∞).
(This question came from an old MATH 100 final exam.)
5. (a) [14] y = f (x) = 8x5 − 5x4 − 20x3 , −∞ < x < ∞.
Critical numbers, intervals where f is increasing or decreasing, local extrema:
f 0 (x) = 40x4 − 20x3 − 60x2 = 20x2 (2x2 − x − 3) = 40x2 (x + 1)(x − 3/2).
f 0 (x) = 0 iff x = −1 or x = 0 or x = 3/2.
Critical numbers are x = −1, 0, 3/2. Note that the tangent lines are horizontal at each of
these critical numbers, because f 0 (x) = 0.
On (−∞, −1): f 0 (x) > 0, f is increasing.
On (−1, 0): f 0 (x) < 0, f is decreasing; f (−1) = 7 is a local maximum value.
On (0, 3/2): f 0 (x) < 0, f is decreasing.
On (3/2, ∞): f 0 (x) > 0, f is increasing; f (3/2) = −513/16 is a local minimum value.
Intervals where the graph is concave upward or downward, and the location of
any inflection points:
f 00 (x) = 160x3 − 60x2 − 120x = 20x(8x2 − 3x − 16).
Possible inflection points (to √be verified below):
3± 201
f 00 (x)
= 0 iff x =0 or x = 16 ≈ −0.7, 1.1.
√
On −∞, 3−16201 : f 00 (x) = 20x(8x2 − 3x − 16) < 0, the graph of f is concave downward.
√
On 3−16201 , 0 : f 00 (x) = 20x(8x2 −3x−16) > 0, the graph of f is concave upward; inflection
√ √
point at x = 3−16201 ≈ −0.7, y = f 3−16201 ≈ 4.3.
√
3+ 201
On 0, 16
: f 00 (x) = 20x(8x2 − 3x − 16) < 0, the graph of f is concave downward;
inflection
√ point at x = 0, y = 0.
On 3+16201 , ∞ : f 00 (x) = 20x(8x2 − 3x − 16) > 0, the graph of f is concave upward;
√ √
3+ 201
inflection point at x = 16 ≈ 1.1, y = f 3+16201 ≈ −20.0.
Behaviour at infinity:
Write f (x) as f (x) = x5 8 − x5 − x202 to see that limx→∞ f (x) = 8 limx→∞ x5 = ∞, and
limx→−∞ f (x) = 8 limx→−∞ x5 = −∞. No asymptotes.
4
Sketch of graph:
(b) [14] y = g(x) = x1/3 − x2/3 , −∞ < x < ∞.
Critical numbers, intervals where g is increasing or decreasing, local extrema:
√
3
x
√
g 0 (x) = 13 x−2/3 − 23 x−1/3 = 1−2x
.
= 23 1/2−
3 2
3x2/3
x
√
g 0 (x) = 0 iff 3 x = 1/2 iff x = 1/8, g 0 (x) does not exist iff x = 0. Critical numbers are
x = 0, 1/8. Note that at x = 1/8 the tangent line is horizontal because g 0 (1/8) = 0, and
at x = 0 the tangent line is vertical because g is continuous at 0, limx→0− g(x) = +∞,
limx→0+ g(x) = +∞.
On (−∞, 0): g 0 (x) > 0, g is increasing.
On (0, 1/8): g 0 (x) > 0, g is increasing (so g(0) is neither a local maximum nor a local
minimum value, and in fact it is correct to say that g is increasing on the open interval
(−∞, 1/8)).
On (1/8, ∞): g 0 (x) < 0, g is decreasing; g(1/8) = 1/4 is a local maximum value.
1/3
Intervals where the graph is concave upward or downward, and the location of
any inflection points:
√
3
x−1
g 00 (x) = − 92 x−5/3 + 29 x−4/3 = 92 √
3 5 .
x
Possible inflection points:
g 00 (x) = 0 iff x = 1, g 00 (x)
does not exist iff x = 0.
√
2 3√x−1
00
On (−∞, 0): g (x) = 9 3 x5 > 0, the graph of g is concave upward.
√
3
x−1
On (0, 1): g 00 (x) = 29 √
3 5 < 0, the graph of g is concave downward; inflection point at x = 0,
x
y = 0.
√
3 x−1
On (1, ∞): g 00 (x) = 29 √
3 5 > 0, the graph of g is concave upward; inflection point at x = 1,
x
y = 0.
Behaviour at infinity:
√
3
1
to see that
Write g(x) as g(x) = x2/3 (x−1/3 − 1) = x2 √
3 x − 1
√
√
3
3
limx→∞ g(x) = − limx→∞ x2 = −∞, and limx→−∞ g(x) = − limx→−∞ x2 = −∞. No
asymptotes.
5
Sketch of graph:
(c) [14] y = h(t) = t2/3 (t − 2)2 , −∞ < t < ∞.
Critical numbers, intervals where h is increasing or decreasing, local extrema:
√
h0 (t) = 38 t5/3 − 20
t2/3 + 83 t−1/3 = 34 t−1/3 (2t2 − 5t + 2) = 38 (t−1/2)(t−2)
.
3
3
t
0
0
h (t) = 0 iff t = 1/2 or t = 2, h (t) does not exist iff t = 0.
Critical numbers are t = 0, 1/2, 2. Note that at t = 1/2 and at t = 2 the tangent
line is horizontal, and at t = 0 the tangent line is vertical because h is continuous at 0,
limt→0− h(t) = −∞, limt→0+ h(t) = +∞,
On (−∞, 0): h0 (t) < 0, h is decreasing.
On (0, 1/2): h0 (t) > 0, h is increasing; h(0) = 0 is a local minimum value.
9
On (1/2, 2): h0 (t) < 0, h is decreasing; h(1/2) = 4 √
is a local maximum value.
3
4
0
On (2, ∞): h (t) > 0, h is increasing; h(2) = 0 is a local minimum value.
Intervals where the graph is concave upward or downward, and the location of
any inflection points:
2
√
t2/3 − 40
t−1/3 − 89 t−4/3 = 98 t−4/3 (5t2 − 5t − 1) = 98 5t −5t−1
.
h00 (t) = 40
3 4
9
9
t
Possible inflection points:
√
h00 (t) = 0 iff 5t2 − 5t − 1 = 0 iff t = 5±10 45 ≈ −0.17, 1.17, h00 (t) does not exist iff t = 0.
√
2
√
On −∞, 5−10 45 : h00 (t) = 98 5t −5t−1
> 0, the graph of h is concave upward.
3 4
t
√
2
√
< 0, the graph of h is concave downward; inflection point
On 5−10 45 , 0 : h00 (t) = 98 5t −5t−1
3 4
t
√ √
at x = 5−10 45 ≈ −0.17, y = h 5−10 45 ≈ 1.45.
√ 2
√
On 0, 5+10 45 : h00 (t) = 89 5t −5t−1
< 0, the graph of h is concave downward (no inflection
3 4
t
√
5− 45
point
at√x = 10 ≈ −0.17).
2
√
> 0, the graph of h is concave upward; inflection point at
On 5+10 45 , ∞ : h00 (t) = 89 5t −5t−1
3
√ t4
√
x = 5+10 45 ≈ 1.17, y = h 5+10 45 ≈ 0.76.
6
Behaviour at infinity:
√
3
Write h(t) as h(t) = t2/3 (t2 − 2t + 4) = t8/3 (1 − 2t−1 + 4t−2 ) = t8 1 − 2t + t42 to see that
√
√
3
3
limt→∞ h(t) = limt→∞ t8 = ∞ and limt→−∞ h(t) = limt→−∞ t8 = ∞. No asymptotes.
Sketch of graph:
7