Math 2250 Written HW #10 Solutions
1. Find the absolute maximum and minimum values of the function f (x) = 9 − |x| subject to
the constraint −1 ≤ x ≤ 3.
Answer: First, remember that we can re-write the absolute value function as a piecewise
function:
(
x
if x ≥ 0
|x| =
−x if x < 0.
Therefore, we can write f (x) as the piecewise function
(
9 − x if x ≥ 0
f (x) =
9 + x if x < 0.
If we differentiate f , then we see that
(
−1
f (x) =
1
0
if x > 0
if x < 0.
In particular, f 0 (x) does not exist at x = 0. Therefore, since f 0 (x) is never 0, x = 0 is the
only critical point of f (x).
Hence, we can just evaluate f (x) at the endpoints and the single critical point:
f (−1) = 9 − | − 1| = 9 − 1 = 8
f (0) = 9 − |0| = 9
f (3) = 9 − |3| = 6
So we see that the absolute maximum value of f (x) on the interval [−1, 3] is 9, which occurs
at x = 0 and the absolute minimum value is 6, which occurs at x = 3.
Here’s the graph of f (x):
8
6
4
2
-1
0
1
1
2
3
2. Find all local maxima and local minima of the function g(x) =
x
.
ex2
Answer: As usual, the first step is to differentiate g:
2
2
x
x
e (1) − x 2xe
g 0 (x) =
2
ex2
g 0 (x) =
0
g (x) =
g 0 (x) =
2
2
2
1 − 2x2
2
e x2
ex − 2x2 ex
2
ex2
ex
1 − 2x2
.
e x2
This is clearly defined everywhere, so the only critical points will occur when g 0 (x) = 0,
meaning when the numerator is zero:
1 − 2x2 = 0
1 = 2x2
r
1
=x
2
1
± √ = x.
2
±
√
So the only two critical points of the function are at x = ±1/ 2 ≈ ±0.707.
√
We can see that g 0 (x) < 0 for x < −1/ 2 by evaluating at x = −1:
1 − 2(−1)2
−1
< 0.
=
2
(−1)
e
e
√
Hence, g(x) is decreasing for −∞ < x < −1/ 2.
√
√
Likewise, we can see that g 0 (x) > 0 for −1/ 2 < x < 1/ 2 by evaluating at x = 0:
g 0 (−1) =
1 − 2(0)2
1
= = 1 > 0.
2
0
1
e
√
√
So g(x) is increasing for −1/ 2 < x < 1/ 2.
√
Finally, we can see that g 0 (x) < 0 for x > 1/ 2 by evaluating at x = 1:
g 0 (0) =
g 0 (1) =
√
So g(x) is decreasing for x > 1/ 2.
−1
1 − 2(1)1
=
< 0.
2
1
e
e
√
Therefore, g(x) switches
√ from decreasing to increasing at x = −1/ 2, so g(x) has a local
√
minimum at x = −1/ 2. Similarly, g(x)
√switches from increasing to decreasing at x = 1/ 2,
so g(x) has a local maximum at x = 1/ 2.
Indeed, this matches what we see on the graph of g(x):
2
1
2ã
- 2
-
1
2
1
2
-
2
1
2ã
3. Determine whether the following statements are true or false. If the statement is true, explain
why. If it is false, give an example of a function which shows that it is false (called a
“counterexample”).
(a) If x = c is a critical point of the function f (x), then it is also a critical point of the
function g(x) = f (x) + k for any constant k.
Answer: True.
Since x = c is a critical point of f (x), we know that either f 0 (c) = 0 or f fails to be
differentiable at x = c.
Now, if g(x) = f (x) + k, then g 0 (x) = f 0 (x). Therefore, if f 0 (c) = 0 then g 0 (c) = 0 as
well, and if f fails to be differentiable at x = c then g fails to be differentiable at x = c
as well.
Either way, x = c is a critical point of g(x).
(b) If x = c is a critical point of the function f (x), then it is also a critical point of the
function h(x) = f (x − k) for any constant k.
Answer: False.
Let f (x) = x2 and let k = 1. We know that f 0 (x) = 2x, so the only critical point of f (x)
is x = 0.
However, h(x) = f (x − 1) = (x − 1)2 , so h0 (x) = 2(x − 1). But then we can see that
h0 (0) = 2(0−1) = −2, so x = 0 is not a critical point of h(x). Indeed, the local minima of
f and h occur at different places in the graph below. This counterexample demonstrates
that the statement cannot possibly be true.
8
6
4
2
-2
-1
1
3
2
3