MATH 1A SECTION: OCTOBER 21, 2013
1. Sketch the graph of a continuous function f on the interval [1, 5] with the following
properties:
(a) absolute minimum at 2, absolute maximum at 3, local minimum at 4.
(b) 2 and 4 are critical numbers, but there are no local maxima or minima.
(c) absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minimum
at 4.
2. Find the absolute minimum and absolute maximum of the function f (x) on the given
interval:
(a) f (x) = 12 + 4x − x2 on the interval [0, 5].
(b) f (x) = x +
1
x
on the interval [0.2, 4].
(c) f (x) = 2 cos x + sin(2x) on the interval [0, π/2].
(d) f (x) = ln(x2 + x + 1) on the interval [−1, 1].
(e) f (x) = x − 2 arctan x on the interval [0, 4].
3. Give an example of a function with 5 as a critical number but which does not have a local
extreme value at x = 5.
4. Show that x101 + x51 + x + 1 has no local maximum nor minimum.
Solutions and Commentary
2. First, we differentiate. Then, set the derivative equal to zero; solving then yields the
critical numbers. Find the value of f (x) at each critical number and each endpoint; the
largest is the absolute maximum, and the smallest is the absolute minimum.
(a) We have f (x) = 12 + 4x − x2 . Then f 0 (x) = 4 − 2x. To find the critical numbers,
we solve 0 = f 0 (x) = 4 − 2x, so 2x = 4 and hence x = 2. The only critical number is
2. Now, we have f (2) = 16, f (0) = 12, and f (5) = 7. So the absolute maximum is
f (2) = 16 and the absolute minimum is f (5) = 7.
(b) We have f (x) = x + x1 . Then f 0 (x) = 1 − x12 . To find the critical numbers, we solve
0 = 1 − x12 , so 1 = x12 , and hence x2 = 1. Therefore, x = ±1, of which only x = 1 is
within our interval. Hence, the critical number is 1.
Now, we have f (1) = 2, f (0.2) = 5.2, and f (4) = 4.25. So the absolute maximum is
f (0.2) = 5.2 and the absolute minimum is f (1) = 2.
(c) We have f (x) = 2 cos x + sin(2x). Then f 0 (x) = −2 sin x + 2 cos(2x). We now solve
0 = f 0 (x) = −2 sin x + 2 cos 2x to find the critical numbers. Then 2 sin x = 2 cos 2x,
so we need to solve sin x = cos 2x.
There are a number of ways to solve this equation, and if you find this difficult, you
may wish to review trigonometry. Here’s one way to do this:
Expand cos 2x using the double angle identity: cos 2x = 1 − 2 sin2 x. Then we are
trying to solve 2 sin x = 1 − 2 sin2 x. This means that 2 sin2 x + 2 sin x − 1 = 0. Let
y = sin x; then this is a quadratic equation 2y 2 + 2y − 1 = 0. We can factor this as
(2y − 1)(y + 1) = 0 to see that sin x = y = 12 and sin x = y = −1. In the interval
[0, π/2], we see that sin x = 21 has the solution π6 , while sin x = −1 has no solutions.
So the only critical number
is π6 .
√
Now, we √have f ( π6 ) = 3 2 3 , f (0) = 2, and f ( π2 ) = 0. So the absolute maximum is
f ( π6 ) = 3 2 3 , and the absolute minimum is f ( π2 ) = 0.
(d) We have f (x) = ln(x2 + x + 1). Then f 0 (x) = x22x+1
. To find critical numbers, solve
+x+1
1
0
f (x) = 0. It suffices to solve 2x + 1 = 0, so x = − 2 . Hence, the only critical number
is − 12 .
Now, we have f (− 12 ) = ln( 34 ), f (0) = 0, and f (1) = ln 3. So we see that the absolute
maximum is f (1) = ln 3 and the absolute minimum is f (− 21 ) = ln( 34 ).
2
0
(e) We have f (x) = x − 2 arctan x. Then f 0 (x) = 1 − 1+x
2 . Solve f (x) = 0 to find
2
2
the critical numbers. That is, we are solving 0 = 1 − 1+x
2 , so 1 = 1+x2 , and hence
1 + x2 = 2, so x2 = 1 and x = ±1. Only x = 1 is within our interval, so our only
critical number is 1.
We now have f (1) = 1 − π2 , f (0) = 0, and f (4) = 4 − 2 arctan 4. Comparing these
values yields that the absolute maximum is f (4) = 4 − 2 arctan(4) and the absolute
minimum is f (1) = 1 − π2 .
3. We want a function f (x) with a critical number of 5 but which does not have a local
extreme value at x = 5. This might remind you of the function g(x) = x3 , which is a
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simple example of a function with a critical number at 0 (since g 0 (x) = 3x2 ) that does not
have a local maximum or minimum at 0.
So we want to translate the function g(x) = x3 rightward by 5 units; this suggests that
f (x) = (x − 5)3 does the job. Indeed, f 0 (x) = 3(x − 5)2 , so f 0 (5) = 0 and 5 is a critical
number, as we expect; you can plot the function (or compute the second derivative) to
see that f (x) does not have a local maximum or minimum at 5; in fact f (x) does not
have any local extreme values anywhere.
4. Consider f (x) = x101 + x51 + x + 1. Then f 0 (x) = 101x100 + 51x50 + 1. Note that x50 and
x100 are both nonnegative quantities (since they are even powers of x). Therefore, we see
that f 0 (x) = 101x100 + 51x50 + 1 ≥ 1. In particular, this means that f 0 (x) 6= 0. Therefore,
f (x) has no critical numbers. This means that f (x) cannot have any local maxima or
minima, which is what we wanted to show.
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