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MA141-008
Test 3 Form A
October 28, 2011
Stephen Adams
North Carolina State University Department of Mathematics
Please put all work in the stamped blue book that has been provided. Nothing on this test
sheet will be graded. Please put one problem on each page; you are allowed to use the back of the
sheet as a new page. No calculators, formula sheets, or other aids are permitted. Please show all
of your work. Simplify all solutions completely and clearly indicate your answers.
1. [12 pts.] Let f (x) = 12 tan x. Use a linear approximation at x = π4 to approximate f (1)
rounded to two decimal places. Use π4 ≈ 0.79 for your approximation.
The we can approximate f (1) by finding the line tangent to f when x = π4 and plugging
x = 1 into the equation of this line. Notice that f ( π4 ) = 12 . f 0 (x) = 21 sec2 x, so f 0 ( π4 ) = 1.
The equation of the tangent line is then y = 1 · (x − π2 ) + 21 . Then f (1) ≈ 1.5 − 0.79 = 0.71.
2. A rectangle has length 20 inches and width 4 inches. The length of the rectangle decreases
at a rate of 2 inches per second while the width of the rectangle increases at a rate of 1 inch
per second.
(a) [12 pts.] At what rate is the area of the rectangle changing after 4 seconds?
(b) [12 pts.] What is the maximum area of the rectangle?
(c) [12 pts.] Show that your answer to part (b) is indeed a maximum.
(a) After x be the length of the rectangle and let y be the width. Then dx
dt = −2 and
dy
dy
dA
dx
dt = 1. The area is given by A = xy and hence dt = y dt + x dt . After four seconds
x = 12 and y = 8. Then dA
dt = 8 · −2 + 12 · 1 = −4. The area is decreasing at a rate of
four square inches per second.
Alternatively, after t seconds the length of the rectangle is 20 − 2t and the width is 4 + t.
Then the area of the rectangle after t seconds is A(t) = (20 − 2t)(4 + t) = 80 + 12t − 2t2 .
dA
Then dA
dt = 12 − 4t. When t = 4, dt = −4.
(b) We have A0 (t) = 12−4t from the above. The maximum area will occur when either A0 (t)
is undefined or A0 (t) = 0. Clearly A0 (t) is defined for all t, so the maximum can only
occur when A0 (t) = 0, and hence when t = 3. The area when t = 3 is A(3) = 14 · 7 = 98
square inches.
(c) A00 (t) = −4 < 0 when t = 3. Then by the Second Derivative Test we see that the area
is indeed maximized when t = 3.
3. (a) [12 pts.] State the Mean Value Theorem.
(b) [12 pts.] Let f be continuous over [1, 3] and differentiable over (1, 3). Suppose that
2 ≤ f 0 (x) ≤ 5 for all x in [1, 3], and f (1) = 4. What is the maximum possible value that
f (3) can be?
MA141-008
Test 3 Form A
October 28, 2011
Stephen Adams
(a) Let f be continuous on [a, b] and differentiable on (a, b). Then there exists a number c
(a)
in (a, b) such that f 0 (c) = f (b)−f
.
b−a
(1)
(b) We know that there is a value of c in (1, 3) such that f 0 (c) = f (3)−f
= f (3)−4
by the
3−1
2
Mean Value Theorem. Then f (3) = 2f 0 (c) + 4. Then the largest possible value of f (3)
will occur when f 0 (c) is a large as possible. This maximum value of f 0 (c) is 5, so the
largest possible value of f (3) is 14.
4. [16 pts.] Let f (x) be a continuous function whose second derivative exists for all real x. f (x)
has the following properties:
• f 0 (x) > 0 on (3, ∞)
• f 0 (x) < 0 on (−∞, 0) ∪ (0, 3)
• f 00 (x) > 0 on (−∞, 0) ∪ (2, ∞)
• f 00 (x) < 0 on (0, 2)
• f (0) = 3, f (2) = −13, and f (3) = −26
• The graph of f (x) has no asymptotes.
Sketch the graph of f .
5. [12 pts.] Evaluate: lim cot (2x) sin (6x).
x→0
Notice that this limit is an indeterminate form of type ∞ · 0. If we can rewrite the limit as
an indeterminate form of type 00 , then we can use L’ôpital’s Rule to evaluate the limit.
sin (6x)
x→0 tan (2x)
6 cos (6x)
= lim
x→0 2 sec2 (2x)
6
=
2
= 3
lim cot (2x) sin (6x) =
x→0
lim