M161 Final Exam Key, Fall 2012
1. (5 points each) Find the following limits if they exist. If the limit equals ±∞ or does not exist, indicate so.
x2 − 4
4−4
=
= 0
2
x→2 2x + 5x + 2
8 + 10 + 2
(a) lim
(b)
x2 − 4
(x + 2)(x − 2)
x−2
−2 − 2
4
= lim
= lim
=
=
2
x→−2 2x + 5x + 2
x→−2 (x + 2)(2x + 1)
x→−2 2x + 1
2(−2) + 1
3
lim
x2 − 4
1
1 − 4/x2
=
=
lim
x→∞ 2x2 + 5x + 2
x→∞ 2 + 5/x + 2/x2
2

x−2

if x < 2,
e
(d) lim f (x) where f (x) = 2
if x = 2,
x→2


x − 1 if x > 2
(c) lim
lim f (x) = lim ex−2 = e2−2 = e0 = 1, and lim f (x) = lim (x − 1 = 2 − 1 = 1. Since these are the same,
x→2−
x→2−
x→2+
x→2+
the limit exists and equals 1
2. (8 points each) Find the derivative of each of the following functions.
(a) f (x) = x ln x
f 0 (x) = 1 · ln x + x ·
√
(b) g(t) =
1
= ln x + 1
x
4t2 + 1
= (4t2 + 1)1/2 e−t ,
et
1
g 0 (t) = (4t2 + 1)−1/2 (8t)e−t + (4t2 + 1)1/2 (−e−t )
2
√
4t2 + 1
1
(c) h(t) = ln
= ln(4t2 + 1) − t
t
e
2
h0 (t) =
1
8t
· 2
−1
2 4t + 1
(d) r(u) = (ln π + e2 )
√
2
The function is a constant independent of u, so r0 (u) = 0.
2
3. (10 points) Using logarithmic differentiation, find the derivative of y = xx .
ln y = x2 ln x, so
y0
1
2
= 2x ln x + x2 · = 2x ln x + x, and y 0 = xx (2x ln x + x).
y
x
4. (10 points) Find an equation of the tangent line to the graph of f (x) =
f 0 (x) = −
2
at the point (2, 1).
x
2
1
, so f 0 (2) = − 12 , and the equation for the tangent line is L(x) = 1 − (x − 2).
2
x
2
2
5. (12 points) Using the following information, sketch the graph of the function f (x) =
be useful to know that 2−1/3 ≈ 0.8.)
2x3
. (It might also
x3 − 1
Domain: (−∞, 1) ∪ (1, ∞)
Intercepts: y-intercept: 0; x-intercept: 0
Asymptotes: Horizontal asymptote y = 2; vertical asymptote x = 1
Intervals where f is increasing and decreasing: Decreasing on (−∞, 1) and on (1, ∞).
Relative extrema: None
Concavity: Downward on (−∞, −2−1/3 ) and (0, 1); upward on (−2−1/3 , 0) and (1, ∞)
Points of inflection: (−2−1/3 , 2/3) and (0, 0).
3 y
2
1
x
−3
−2
−1
1
2
3
−1
−2
−3
6. (10 points) Use implicit differentiation to find
dy
for yex = 3x + y 2 .
dx
dy x
dy
dy
3 − yex
e + yex = 3 + 2y , so
= x
.
dx
dx
dx
e − 2y
7. (12 points) Find the absolute maximum value and the absolute minimum value, if any, of f (x) = x3 + 6x2 + 2
on [−2, 1].
f 0 (x) = 3x2 +12x = 3x(x+4) has roots x1 = −4 and x2 = 0. Since −4 is not in [−2, 1], we only have to evaluate
f (x) at the endpoints and at x2 = 0 and compare values. We get f (−2) = (−2)3 +6(−2)2 +2 = −8+24+2 = 18,
f (1) = 1 + 6 + 2 = 9, and f (0) = 2, so the maximum is 18 and the minimum is 2.
8. (10 points each) Find the following indefinite integrals.
Z
(a)
Z
(b)
√ x − x dx =
2
2
3
3
Z 3x (x − 3) dx =
Z
x3 2 3/2
x2 − x1/2 dx =
− x +C
3
3
u3 du =
u4
(x3 − 3)4
+C =
+ C, with substitution u = x3 − 3, du = 3x2 dx.
4
4
3
√
(c)
(e + 5 − t) dt = e dt +
5 − t dt. The first integral with the substitution u = 2t, du = 2 dt, is
Z
Z
1
1
1
e2t dt =
eu du = eu + C1 = e2t + C1 , the second integral with the substitution v = 5 − t, dv = − dt, is
2
2
2
Z
Z
Z
√
√
√
2 3/2
2
5 − t dt = −
v dv = − v + C2 = − (5 − t)3/2 . Combining those results gives
(e2t + 5 − t) dt =
3
3
1 2t 2
e − (5 − t)3/2 + C, where we combined the constants as C = C1 + C2 .
2
3
Z
2t
√
Z
2t
Z
Z
s2 + 1
1
du
1
1
(d)
ds =
= ln |u| + C =
ln |s3 + 3s + 1| + C, where we used the substitution
3
s + 3s + 1
3
u
3
3
u = s3 + 3s + 1, du = (3s2 + 3) ds = 3(s2 + 1) ds
Z 2
Z 2
2
2x + x2 + 3x4
2
2
9. (10 points) Find
dx =
+ 1 + 3x
dx = 2 ln x + x + x3 1 = 2 ln 2 + 2 + 23 − (ln 1 +
2
x
x
1
1
1 + 13 ) = 2 ln 2 + 8
Z
10. (10 points) Find the average value of f (x) =
1
Average is
e−1
Z
1
e
1
on the interval [1, e].
x
1
1
1
1
1
dx =
[ln |x|]e1 =
(ln e − ln 1) =
(1 − 0) =
.
x
e−1
e−1
e−1
e−1
11. (12 points) Find the area of the region completely enclosed by the graphs of f (x) = x2 −5 and g(x) = 3−x2 .
Points of intersection: x2 − 5 = 3 − x2 is equivalent to 2x2 = 8, which has solutions x1 = −2 and x2 = 2.
Since f (0) = −5 and g(0) = 3, and both functions are continuous, we have that f (x) ≤ g(x) for −2 ≤ x ≤ 2,
Z 2
Z 2
Z 2
2 3 2
2
2
2
=
and so the area is
(g(x) − f (x)) dx =
((3 − x ) − (x − 5)) dx =
(8 − 2x ) dx = 8x − x
3
−2
−2
−2
−2
2
2
64
8 · 2 − · 23 − 8 · (−2) − · (−2)3 =
.
3
3
3
12. (10 points) A boat is pulled toward a dock by means of a rope wound on a drum that is located 5 ft above
the bow of the boat. If the rope is being pulled in at the rate of 2 ft/sec, how fast is the boat approaching the
dock when it is 12 ft from the dock. (Hint: 52 + 122 = 132 .)
(We suppress the units here, all length are in feet and all time derivatives are feet per second.) Let A denote
the distance of the boat to the dock, B the height of the dock, and C the length of the rope. By Pythagoras’s
dB
dC
dA
theorem, A2 + B 2 = C 2 , and differentiating this gives 2A ·
+ 2B ·
= 2C ·
. Since B = 5 is constant,
dt
dt
dt
√
√
dB
dC
we get
= 0. The other conditions give
= −2, A = 12 and C = A2 + B 2 = 122 + 52 = 13 at the
dt
dt
dA
dA
2 · 13 · (−2)
13
specified time t. Plugging this into the equation gives 2·12·
= 2·13·(−2), and so
=
=− .
dt
dt
2 · 12
6
So the boat is approaching the dock at a speed of 13/6 ft/sec.
4
13. (12 points) A storage shed with a square base and no floor, with a square flat roof, is to have a volume of
750 cubic feet. If the material for the roof costs $3 per square foot, and the material for the sides cost $2 per
square foot, determine the dimensions of the shed that can be constructed at minimum cost.
The cost for the roof is 3x2 , for the four sides it is 4 · 2xh, so the total cost is 3x2 + 8xh. The volume is
x2 h = 750, so h = 750/x2 , and the total cost in terms of x is f (x) = 3x2 + 8x(750/x2 ) = 3x2 + 6000x−1 . To
find the critical points, we solve f 0 (x)
= 0, so 6x − 6000x−2 = 0. Multiplying through with x2 and dividing
√
by 6 gives x3 − 1000 = 0, so x = 3 1000 = 10, and h = 750/x2 = 750/100 = 7.5. To check whether the
critical point is really a minimum, we find f 00 (x) = 6 + 12000x−3 > 0, which shows that the function is concave
up, and so the critical point is really the unique minimum. So the dimensions of the box of minimal cost are
x = 10 feet and h = 7.5 feet.