Exam 3 Solutions
1. (20 points) Consider the function
f (x) = 3x4 − 4x3 + 43
a. Find any/all critical points.
b. Find any/all local max/mins.
c. Find any/all inflection points.
We start by taking the first derivative:
f 0 (x) = 12x3 − 12x2
c is a criticial point of f
⇐⇒
((((
f 0 (c)
undefined
f 0 (c) = 0 or (
((
⇐⇒
12c3 − 12c2 = 0
⇐⇒
12c2 (c − 1) = 0
⇐⇒
c = 0, 1
Making a sign chart for f 0 ,
we see f has no local maxima and one local minimum at x = 1.
1
Now, we analyze the second derivative:
f 00 (x) = 36x2 − 24x
(((( ⇐⇒
f 00 (c) = 0 or (
f 00(
(c)(undefined
⇐⇒
12c(3c − 2) = 0
⇐⇒
c = 0,
Making a sign chart for f 00 ,
we see f has two inflection points at x = 0, 23 .
2
36c2 − 24c = 0
2
3
2. (20 points) Consider the function
f (x) =
x2
(x − 3)2
a. Find any/all critical points.
b. Find any/all local max/mins.
c. Find any/all inflection points.
Again, start by taking the first derivative:
f 0 (x)
=
=
x2
(x − 3)2
0
2x · (x − 3)2 − x2 · 2(x − 3)
(x − 3)4
2x · (x − 3) (x − 3) − x
=
=
(x − 3)4
−6x
(x − 3)3
x is a criticial point of f
⇐⇒
f 0 (x) = 0 or f 0 (x) undefined
⇐⇒
−6x = 0 or (x − 3)4 = 0
⇐⇒
x = 0 or x = 3
3
Making a sign chart for f 0 ,
Thus f has a local minimum at x = 0. While it may look like f has a local maximum at x = 3, notice that f is not defined there; (f has a vertical
asymptote at x = 3.)
Now, we analyze the second derivative:
f 00 (x)
=
=
−6x
(x − 3)3
0
(−6) · (x − 3)3 − (−6x) · 3 · (x − 3)2
(x − 3)6
−6 · (x − 3)2 (x − 3) − 3x
=
(x − 3)6
=
−6(−2x − 3)
(x − 3)4
=
6(2x + 3)
(x − 3)4
f 00 (x) = 0 or f 00 (x) undefined ⇐⇒
⇐⇒
4
6(2x + 3) = 0 or (x − 3)4 = 0
x=−
3
or x = 3
2
Making a sign chart for f 00 ,
we see f has only one inflection point at x = − 23 .
3. (10 points) Use the second derivative test (or any method for partial credit)
to find the local max/mins, if any, of
f (x) = x +
0
f (x)
x is a criticial point of f
9
x
=
0
−1
x + 9x
=
1 − 9x−2
=
1−
=
x2 − 9
x2
9
x2
⇐⇒
f 0 (x) = 0 or f 0 (x) undefined
⇐⇒
x2 − 9 = 0 or x2 = 0
⇐⇒
x=
0
x = ±3 or We throw x = 0 away as it is not in the domain of f and hence not a candidate
for a local max/min. Thus, we plug x = 3, −3 into f 00 .
5
f 00 (x) =
1 − 9x−2
0
= 18x−3 =
18
x3
The second derivative test applied to the critical points x = 3, −3:
•
f 00 (3) =
•
f 00 (−3) =
18
(3)3
> 0 =⇒ f has a local minimum at x = 3.
18
(−3)3
< 0 =⇒ f has a local maximum at x = −3.
Reminder: The way to remember this is picturing a ’typical’ (’smooth’) local
max, which is concave down, and vice versa.
4. (10 points) Find the absolute maximum and absolute minimum of
f (x) = x3 − 27x − 12 on [−4, 10]
Recall: The absolute extrema happen either at the endpoints or at interior
critical points.
As f 0 (x) = 3x2 − 27, f has two critical points, x = 3, −3, both of which are
included in (the interior of) the interval.
• f (−4) = (−4)3 − 27(−4) − 12 = 32
• f (−3) = (−3)3 − 27(−3) − 12 = 42
• f (3) = (3)3 − 27(3) − 12 = −66
• f (10) = (10)3 − 27(10) − 12 = 718
Thus, the absolute maximum is 718; The absolute minimum is -66.
6
5. (20 points) You want to enclose three adjacent rectangular pens (see board
for picture). If you have 200 yards of fencing, what is the maximum total area
(in square yards) you can enclose?
Here’s the (correct!) picture:
We want to maximize the total area, A:
A = lw
To do this, we will use all of the fencing:
2l + 4w = 200
This means, for example, once we choose the width, the length is determined:
l = 100 − 2w
This allows us to write the total area as function of just the one variable w:
A = 100 − 2w w = 100w − 2w2
where 0 ≤ w ≤ 50. (The upper bound corresponds to l = 0.)
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Now,
A0 = 100 − 4w
Thus, the only critical point is w = 25.
Of course, A = 0 at the endpoints w = 0 and w = 50 (where l = 0), so
the absolute maximum total area is
Amax = 100 − 2(25) (25) = 50 · 25 = 1, 250 yd2
8
6. (20 points) You are going to build an open box by cutting away equal squares
from the corners of an 8-inch-by-8-inch square of cardboard, then folding up the
resulting ’flaps’. (See board for picture.) What is the largest volume (in cubic
inches) of such a box?
Here’s the picture, including the length and width of the resulting box:
Thus, the volume of the box, V , is given by
V = x(8 − 2x)2
where 0 ≤ x ≤ 4.
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Taking the derivative of V , we have:
0
V (x)
=
0
2
x · (8 − 2x)
=
1 · (8 − 2x)2 + x · 2 · (8 − 2x) · (−2)
=
(8 − 2x)2 − 4x(8 − 2x)
=
(8 − 2x)[(8 − 2x) − 4x]
=
(8 − 2x)(8 − 6x)
Thus, the critical points are x = 4 and x = 4/3.
Again, V = 0 at the endpoints so the absolute maximum is achieved at the
only interior criticial point x = 4/3:
Vmax =
2 2
4
4
4
16
· 8−2·
=
·
≈ 37.9 in3
3
3
3
3
10