Math 112 (Calculus I)
Final Exam KEY
Short Answer Fill in the blank with the appropriate answer.
1. (10 points)
a)
d
sec2 x
ln(tan x) =
dx
tan x
b) Use the linearization of f (x) = x1/3 at a = 8 to approximate 91/3 .
c) If f 0 (x) = x3 and f (0) = 5 then f (x) =
25
12
x4
+ 5.
4
d) If f (x) = e2x then f 00 (x) = 4e2x .
e) The Mean Value Theorem says that if f is a continuous function on [a, b]
which is also differentiable on (a, b) then there is a c ∈ (a, b)
with f 0 (c) =
f (b) − f (a)
.
b−a
f) Circle the correct answer in both cases: If f 0 is positive and increasing, then f is
( increasing / decreasing) and ( concave up / concave down).
x3 +5
3
x→∞ 2x +3x+2
g) lim
=
1
.
2
Multiple Choice. In the grid below fill in the square corresponding to each correct
answer.
2
A
B
C
D
E
F
G
H
I
J
3
A
B
C
D
E
F
G
H
I
J
4
A
B
C
D
E
F
G
H
I
J
5
A
B
C
D
E
F
G
H
I
J
6
A
B
C
D
E
F
G
H
I
J
7
A
B
C
D
E
F
G
H
I
J
8
A
B
C
D
E
F
G
H
I
J
9
A
B
C
D
E
F
G
H
I
J
10
A
B
C
D
E
F
G
H
I
J
2. If sin(xy) =
x−1
, what is y 0 at (1, π)?
y
a) π
d)
−π 2 − 1
π
b) π 2 − 1
e) −1 −
1
π
c) π −
f) −
1
π
1
π
3. Which of the following has a removable discontinuity at x = 1?
a)
x2 − 1
x+2
b)
ln |x2 − 1|
c)
x2 + 2
x−1
d)
x2 − 3x + 2
x2 − 1
e)
tan
πx
2
f)
sin
1
1−x
a)
1
t2 − t
, at what value(s) of t is the velocity 0?
t−1
√
√
b) 1 ± 2
c) 2 ± 2
d)
√
1−2 2
e)
4. If the position of a particle is given by
0
f)
The velocity is never 0.
5. If f (x) is a differentiable function, which of the following is not always true?
a)
Z x
b)
F (t)dt = F 0 (x) − F 0 (a)
e) ZIf F 0 (x) = f (x) then
x
f (t)dt = F (x) − F (a)
a
d)
Z x
d Zx
f (t)dt = f (x)
dx a
f 0 (t)dt = f (x) − f (a)
a
c)
Z
f 0 (x)dx = f (x) + C
a
6. A 20 ft ladder is placed against a wall. If the top of the ladder is dropping at a rate of 5
ft/min, how fast is the bottom of the ladder moving away from the wall (in feet per minute)
when the top of the ladder is 12 ft high?
a)
-32
b)
-160
c)
160
d)
15
2
e)
15
4
f)
15
16
7. If h(x) = x cosh(x2 − 4), what is h00 (2)?
a)
0
b)
4
c)
8
d)
16
e)
32
f)
48
8. Let F (x) =
Z 2x2
3
sin(t)
dt be defined for x > −1. Find F 0 (x).
3
t +1
a)
sin(2x2 )
8x3 + 1
b)
4x sin(2x2 )
8x6 + 1
c)
sin(2x2 ) sin(18)
−
8x3 + 1
216
d)
4x sin(2x2 ) 12 sin(3)
−
8x3 + 1
28
e)
sin(x)
x3 + 1
f)
sin(x) sin(3)
−
x3 + 1
28
Solution: b)
9. If f (x) is a continuous function on the interval [4, 5] and if f (4) = 3 and f (5) = 1, then which
theorem guarantees that there is some value c ∈ [4, 5] such that f (c) = 2?
a)
Mean value Theorem,
b)
Extreme Value Theorem,
c)
Intermediate Value Theorem
d)
The Fundamental Theorem of Calculus
e)
Rolle’s Theorem
f)
No theorem guarantees this because it is
false.
Solution: c)
10. Use one iteration of Newton’s method, beginning with x1 = √
1/2 to approximate the positive
2
root of the equation x + 2x − 1 = 0. (Note that the root is 2 − 1).
a)
1
12
b)
5
12
c)
d)
1
2
e)
7
12
f) x2 is undefined.
0
Free Response. For problems 11 - 19, write your answers in the space provided.
Use the back of the page if needed, indicating that fact. Neatly show all work.
11. (7 points) State the definition of the derivative of f (t), and use the definition to find the
derivative of f (t) = 2t2 + 1.
Solution:
f (x + h) − f (h)
.
h→0
h
f 0 (x) = lim
The other definition is ok as well.
2(t + h)2 + 1 − (2t2 + 1)
h→0
h
f 0 (t) = lim
2t2 + 4th + 2h2 − 2t2
= lim (4t + h) = 4t.
h→0
h→0
h
√
12. (7 points) Find the equation of the tangent line to y = x x + 1 at the point (3, 6).
= lim
Solution:
√
x
x+1+ √
2 x+1
11
3
y 0 (3) = 2 + =
4
4
11
y − 6 = (x − 3)
4
11
9
y = x−
4
4
y0 =
Z 3
x
dx.
+1
1
Solution: Let u = 2x2 + 1. Then du = 4x dx. The integral becomes
13. (7 points) Find the integral
2x2
Z 19
3
ln(19) − ln(3)
1
1
du = ln u|13 9 =
.
4u
4
4
d cos x
x
=
dx
Solution: If y = xcos x , then ln y = cos x ln x. Differentiating, we have
14. (7 points)
1 0
cos x
y = − sin x ln x +
.
y
x
Thus,
y0 = (
cos x
− sin x ln x)xcos x .
x
15. (7 points) Find the absolute minimum and maximum of f (x) = x2 e−x
2 /2
on [0, 4].
Solution:
f 0 (x) = 2xe−x
2 /2
− x3 e−x
2 /2
= x(2 − x2 )e−x
2 /2
.
√
Notice that there are two√critical points of interest in this interval, 0 and 2. Also,
√ it is clear
that on the interval (0, 2) f 0 is positive, so f is increasing. On the interval, ( √2, 4) f 0 is
negative, so f is decreasing. Hence, the maximum value of the function occurs at 2 and is
2
.
e
The minimum must occur at the end points. The smallest value occurs at 0, so the minimum
is 0.
(1 − x)ex − 1
16. (7 points) Find lim
.
x→0
x2
Solution:
−ex + (1 − x)ex
−ex − ex + (1 − x)ex
−1
(1 − x)ex − 1
=
lim
=
lim
=
.
2
x→0
x→0
x→0
x
2x
2
2
lim
17. (7 points) An open-at-the-top vertical tank has horizontal cross-section of an equilateral triangle, and volume of 4000 cubic feet. Find the
√ dimensions that minimize the surface area.
3
(Hint: the height of an equilateral triangle is
times the length of a side.)
2
Solution: The area of the equilateral base is
√
√
1 3
3 2
s·s=
s.
2 2
4
Thus, the volume of the tank is
√
3 2
s h = 4000.
4
Solving for h, we have
16000
h= √ 2.
3s
The surface area of the tank is given by
√
√
3 2
3 2
16000
S=
s + 3sh =
s + 3s · √ 2
4
4
3s
√
√
3 2 16000 3
=
s +
.
4
s
Thus,
√
√
3
16000
3 √ s3 − 32000
S0 =
s−
= 3
.
2
s2
2s2
√
16000
40
Setting S 0 = 0 we have s3 = 32000 or s = 20 3 4. Hence, h = √
= √ 2/3 .
2/3
3400 · 4
34
18. (7 points)Given the following graph for f (x), sketch f 0 .
19. (7 points) Prove that x→a
lim 2x = 2a, using the , δ definition of limit.
Solution: Let > 0 be given. Set δ = . Thus, if
2
0 < |x − a| < δ,
2
⇒ 2|x − a| < |x − a| <
or
|2x − 2a| < .
By the definition of limit, we are done.
END OF EXAM