MATH 131:100 Exam #2
Instructor: Keaton Hamm
June 19, 2014
Print Last Name: SOLUTION
First Name: SOLUTION
Signature: S OL U T I ON
“An Aggie does not lie, cheat or steal or tolerate those who do.”
Instructions:
ˆYou MUST clear all memory on your calculator: to do this hit 2nd, then +
(MEM), then Reset (7), move the cursor over to the right ALL, select All Memory
(1), then Reset (2)
ˆShow your work for each work out problem clearly and legibly. Box your final
answer in the work out problems.
ˆPoint values are shown for each problem.
ˆYou may use a TI-83, TI-84, or TI-Nspire nonCAS version for all problems. Other
calculators are not permitted.
ˆIf you need extra scratch paper, ask me and I will provide you with some.
ˆYour grade will be written on the final page of the exam.
ˆThere should be 8 pages with problems numbered 1-18 on this exam. There are
100 possible points.
Good Luck!
Section 1: Definitions
Give the mathematical definition for each of the following terms or concepts. (Each
problem is worth 3 Points)
1) Provide one of the limit definitions for the derivative of f at the point a:
f (x) − f (a)
x→a
x−a
lim
2) Provide the other limit definition for the derivative of f at the point a:
f (a + h) − f (a)
h→0
h
lim
3) Suppose f and g are differentiable. Then the Product Rule states:
d
[f (x)g(x)] = f (x)g 0 (x) + f 0 (x)g(x)
dx
4) Suppose f and g are differentiable. Then the Quotient Rule states:
d f (x)
g(x)f 0 (x) − f (x)g 0 (x)
=
dx g(x)
(g(x))2
5) Suppose f and g are differentiable. Then the Chain Rule states:
d
f (g(x)) = f 0 (g(x)) · g 0 (x)
dx
Section 2: Graphical Concepts
6) (4 Points) Sketch a graph of a function f such that f 0 is always positive and f 00
is always negative.
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NOTE: Since f 00 is always negative, the function must always be concave down, and
can never change concavity. Likewise, its slope must always be positive, so the function
must be always increasing.
7) (15 Points, 3 Each) The following is the graph of a function f (x).
6
5
4
3
2
1
−2
−1
1
−1
2
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4
5
6
7
8
9
10
11 12
13
−2
−3
−4
a) Where is f not differentiable? x = 4, 9
b) Where is f 0 (x) = 0? x = 0
c) Where is f increasing? (−∞, 0) ∪ (4, 9) ∪ (9, ∞)
d) Where is f decreasing? (0, 4)
e) Where does f attain a local maximum? x = 0
8) (4 Points) Sketch a graph of a function that is continuous everywhere, but not
differentiable at x = 2.
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9) (14 Points, 2 Each) The following is the graph of f 0 (x).
6
5
4
3
2
1
−2
−1
1
−1
2
3
4
5
6
7
8
9
10
11 12
−2
−3
−4
a) Where does f attain a local maximum? x = 4
b) Where does f attain a local minimum? x = 2, 6
c) Where is f increasing? (2, 4) ∪ (6, ∞)
d) Where is f decreasing? (−∞, 2) ∪ (4, 6)
e) Where is f concave up? (−∞, 3) ∪ (5, 7) ∪ (8, ∞)
f ) Where is f concave down? (3, 5) ∪ (7, 8)
g) Where are the inflection points of f , if any? x = 3, 5, 7, 8
13
Section 3: Work Out Problems
10) (5 Points) Supposing that f and g are differentiable functions, differentiate
F (x) = (f (x)g(x))6
SOLUTION: Using the Chain Rule with the outside function being (·)6 and the
inside function being f (x)g(x), we get
F 0 (x) = 6 (f (x)g(x))5
d
(f (x)g(x))
dx
By the Product Rule,
d
(f (x)g(x)) = f (x)g 0 (x) + f 0 (x)g(x)
dx
So
F 0 (x) = 6 (f (x)g(x))5 (f (x)g 0 (x) + f 0 (x)g(x))
11) (5 Points) Suppose that f 0 (6) = 7 and g(x) = x2 + 2. If F (x) = f (g(x)), then
what is F 0 (2)?
SOLUTION: By the Chain Rule,
F 0 (x) = f 0 (g(x)) · g 0 (x)
so
F 0 (2) = f 0 (g(2)) · g 0 (2).
Now g 0 (x) = 2x, so g 0 (2) = 4. Also, g(2) = 22 + 2 = 6. So f 0 (g(2)) = g 0 (6) = 7 as
given by the problem statement. Therefore
F 0 (2) = 7 · 4 = 28
12) (4 Points) Differentiate
g(x) =
√
3
x4 − 7
SOLUTION: Rewrite as a power:
1
g(x) = x4 − 7 3
Then by the Chain Rule with the fact that
d
(x4
dx
− 7) = 4x3 ,
2
1
g 0 (x) = (x4 − 7)− 3 4x3
3
13) (6 Points) Differentiate
f (x) = ln (cos (ex ))
SOLUTION: By the Chain Rule,
f 0 (x) =
1
d
·
(cos(ex )) .
x
cos(e ) dx
Again by the Chain Rule,
d
d
(cos(ex )) = − sin(ex ) (ex ) = − sin(ex )ex .
dx
dx
Then
f 0 (x) =
1
(− sin(ex )ex ) = − tan(ex )ex .
cos(ex )
14) (5 Points) Differentiate (no need to simplify)
y=
x
1 + x2
SOLUTION: By the Quotient Rule:
y0 =
d
d
(1 + x2 ) dx
(x) − x dx
(1 + x2 )
1 + x2 − 2x2
=
(1 + x2 )2
(1 + x2 )2
17) (5 Points) Differentiate
f (x) = eπx x4 + 2x
SOLUTION: Using the Product Rule:
f 0 (x) = eπx
d 4
d
(x + 2x ) + (x4 + 2x ) (eπx )
dx
dx
= eπx (4x3 + 2x ln 2) + (x4 + 2x ) πeπx
16) (3 Points) Evaluate
x1000 − 1
x→1 x − 1
(Hint: This should look like the derivative of something)
lim
SOLUTION: As the hint suggests, this should remind you of the formula
f (x) − f (1)
.
x→1
x−1
f 0 (1) = lim
This would suggest that we define f (x) = x1000 , which makes sense because f (1) =
11000 = 1. Therefore, since f 0 (x) = 1000x999 , the limit is
f 0 (1) = 1000 · 1999 = 1000
15) (6 Points) Find the equation of the tangent line to the curve y = x3 +6x2 −3x−1
at the point (2, 25).
SOLUTION: First find the slope by computing the derivative
y 0 = 3x2 + 12x − 3
and
y 0 (2) = 3(4) + 12(2) − 3 = 33
Using Point-Slope Form, the equation of the tangent line is
y − 25 = 33(x − 2)
or
y = 33x − 41
18) (9 Points) The Surface Area of a sphere is given by the formula S = 4πr2 .
a) Find a formula for the rate of increase of S with respect to r.
SOLUTION:
dS
= S 0 (r) = 8πr
dr
b) What is the rate of increase at r = 2?
SOLUTION:
S 0 (2) = 8π(2) = 16π
c) Find the linearization of S about the point r = 2.
SOLUTION: The formula is
L(r) = S(2) + S 0 (2)(r − 2)
S(2) = 4π(22 ) = 16π. So
L(r) = 16π + 16π(r − 2) = 16πr − 16π
d) Use your answer from part (c) to estimate the Surface Area when r = 3.
SOLUTION:
L(3) = 16π(3) − 16π = 32π
e) Compare your answer from part (d) with the real answer (by evaluating S(3)).
SOLUTION:
S(3) = 2π(32 ) = 36π
So the Linearization was an underestimate. The difference in the actual and estimated
values is 4π.
Final Score: 100