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Math 105: Review for Final Exam, Part II - SOLUTIONS
- Consider the function f(x) = x
3
f(x) = x ln x
3
f(x) = xln x
3
ln x on the interval [1/e, e
2
[1/e, e]
2
[1/e, e]
2
].
(a) Find the xxx- and yyy-coordinates of any and all local extrema and classify each as a local
maximum or local minimum.
f
′
(x) = 3x
2
ln x + x
3
x
0 = x
2
(3 ln x + 1)
⇒ x
2
= 0 (not in our domain) or ln x = − 1 / 3 , which means x = e
− 1 / 3
0 < x < e
− 1 / 3
e
− 1 / 3
< x
f
′
negative positive
f ↘ ↗
y-value: f(e
− 1 / 3
) = (e
− 1 / 3
3
ln(e
− 1 / 3
) = (e
− 1
)(− 1 /3) = (− 1 / 3 e)
So, f has a local minimum at (e
− 1 / 3
, − 1 / 3 e).
(b) Find the x
x x- and y
y y-coordinates of any and all global extrema and classify each as a
global maximum or global minimum.
We check the y-values at the local extrema and the endpoints.
f(1/e) = (1/e)
3
ln(1/e) = (1/e
3
)(−1) = (− 1 /e
3
f(e
− 1 / 3
) = (− 1 / 3 e) from above
f(e
2
) = (e
2
3
ln(e
2
) = (e
6
)(2) = 2e
6
So, f has a global minimum at (e
− 1 / 3
, − 1 /3) and a global maximum at (e
2
, 2 e
6
(c) Find the x
x x-coordinate(s) of any and all inflection points.
f
′′
(x) = 2x(3 ln x + 1) + x
2
x
0 = 6x ln x + 2x + 3x
0 = x(6 ln x + 5)
⇒ x = 0 (not in our domain) or ln x = − 5 / 6 , which means x = e
− 5 / 6
0 < x < e
− 5 / 6
e
− 5 / 6
< x
f
′′
negative positive
f concave down concave up
So, the x-value of the inflection point of f is x = e
− 5 / 6
- Your company is mass-producing a cylindrical container. The flat portion (top and bot-
tom) costs 3 cents per square inch and the curved (lateral) portion costs 5 cents per
square inch. If your budget is $9.00 per container, what dimensions will give the largest
volume?
area of circle = πr
2
πr
2
πr
2
lateral area of cylinder = 222 πrhπrhπrh volume of cylinder = πr
2
πrh
2
πrh
2
h
Objective function: volume = V = πr
2
h
We need to get this down to a function of just one variable, so we use the
constraint equation : cost = 900 = 3 · 2 · πr
2
900 = 6πr
2
900 − 6 πr
2
= 10πrh
900 − 6 πr
2
10 πr
= h
Substituting this back into the objective function gives
V = πr
2
h = πr
2
900 − 6 πr
2
10 πr
= r ·
900 − 6 πr
2
(900r − 6 πr
3
Now that we have V as a function of just one variable, we find its maximum.
V
′
(x) =
(900 − 18 πr
2
(900 − 18 πr
2
⇒ 18 πr
2
⇒ r
2
π
⇒ r =
π
0 < x <
50 /π
50 /π < x
f
′
positive negative
f ↗ ↘
Thus, we have in fact found the global maximum at r =
50 /π.
And h =
900 − 6 πr
2
10 πr
= ...much simplifying... =
π
≈ 4 .787 inches.
- You are standing on a pier, 6 feet above the deck of a boat. Attached to the boat is a
line, which you are pulling in at a rate of 3 feet per second. When there are 10 feet of
line between your hand and the boat, at what rate is the boat moving across the water?
a
b
You
Boat
We know
db
dt
, and we want to find
da
dt
So, we write an equation that relates a and b and then differentiate implicitly with respect to time t.
a
2
2
= b
2
2 a
da
dt
db
dt
da
dt
b
a
db
dt
At the moment in question, b = 10, a = 8 (by the Pythagorean Theorem), and
db
dt
So,
da
dt
· (−3) = − 3 .75 feet per second, meaning the boat is moving toward the dock at 3.75 feet
per second.
- Use the Intermediate Value Theorem to show that f(x) = x
3
− 2 x − 1
f(x) = x
3
− 2 x − 1 f(x) = x
3
− 2 x − 1 has a root on [1, 2]
[1, 2]
[1, 2].
IVT: If f is continuous on [a, b] and y is a number between f(a) and f(b), then there is a number c
between a and b such that f(c) = y.
For the function given above, f(1) = −2 and f(2) = 3. Since 0 is a number between −2 and 3, the
IVT says there is a number c between 1 and 2 such that f(c) = 0; this c is the desired root.
Finally, we are taking the sine of each of our x-values, so the function in question must be
f(x) = sin x.
Thus, we see that the expression is equal to the area under f(x) = sin x on [0, π].
Its value is
π
0
sin x dx = − cos x
π
0
= − cos(π) − (− cos(0)) = −(−1) − (−1) = 2.
- Water is leaking out of a tank at a decreasing rate rrr(((ttt))) as shown below.
time (min) 0
rate (gal/min) 151515 111111 888 444 333
(a) Find an overestimate and underestimate for the total amount that leaked out during
these 8 minutes.
overestimate = L 4
underestimate = R 4
(b) Interpret the expression
6
2
r(t) dt
6
2
r(t) dt
6
2
r(t) dt in terms of the situation described above.
This integral gives the amount (in gallons) of water that leaked from the tank on the interval
[2, 6] minutes.
- Consider the graph of fff(((ttt))) shown. It is made of straight lines and a semicircle.
-4 -3 -2 -1 0 1 2 3 4
2
1
0
t
f(t)
Let G(x) =
x
0
G(x) = f(t) dt
x
0
G(x) = f(t) dt
x
0
f(t) dt and H(x) =
x
− 3
H(x) = f(t) dt
x
− 3
H(x) = f(t) dt
x
− 3
f(t) dt.
(a) Compute GGG(2)(2)(2), GGG(4)(4)(4), and HHH(4)(4)(4).
G(2) is the area under f between t = 0 and t = 2. This is a rectangle plus a triangle and has area
Similarly, G(4) = 2 · 1 +
π(1)
2
π
H(4) is the area under f between t = −3 and t = 4. Remember that area below the t-axis counts
as negative.
H(4) = − (2 · 1 +
· 2 · 1 + [area under f from 0 to 4, found above as G(4)]
[
π
]
π
(b) Where is G
G
G increasing? Where is G
G
G decreasing?
G is increasing where f is positive: (− 1 , 4]. Note that G has a horizontal slope at x = 2 but since
f is positive on each side of t = 2, we say G is increasing at x = 2.
G is decreasing where f is negative: [− 4 , −1).
(c) Where is G
G
G concave up? Where is G
G
G concave down?
G is concave up where f is increasing: (− 2 , 0) ∪ (2, 3).
G is concave down where f is decreasing: (1, 2) ∪ (3, 4].
(d) At what x
x x-value(s) does G
G
G have a local maximum? At what x
x x-value(s) does G
G
G have a
local minimum?
G has a local maximum where f changes from positive to negative: never.
G has a local minimum where f changes from negative to positive: x = −1.
(e) Find a formula that relates GGG and HHH.
From their definitions, H(x) =
0
− 3
f(t) dt + G(x) = −2 + G(x).
(f) How would your answers to (b), (c), and (d) change if the questions were about H
H
H
instead of G
G
G?
They would not change at all because H
′
(x) = G
′
(x).
- (a) Use sigma notation to express L 10
L
10
L
10
and M
10
M
10
M
10
as approximations to
60
20
ln x dx
60
20
ln x dx
60
20
ln x dx.
We’re subdividing the interval into 10 pieces, so each piece has width ∆x =
L
10
= [f(20) + f(24) + f(28) + ... + f(52) + f(56)]∆x
= [ln(20) + ln(24) + ln(28) + ... + ln(52) + ln(56)] · 4
9
k=
ln(20 + 4k) · 4
M
10
= [f(22) + f(26) + f(30) + ... + f(54) + f(58)]∆x
= [ln(22) + ln(26) + ln(30) + ... + ln(54) + ln(58)] · 4
9
k=
ln(22 + 4k) · 4
(b) Draw a sketch that represents the sum M
4
M
4
M
4
Now we’re subdividing the interval into 4 pieces, so each piece has width ∆x =
Note that the height of each rectangle is determined by the y-value of the curve at the middle
x-value of the rectangle (that is, at x = 25, 35, 45, 55).
y=ln x
