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Math 105: Review for Final Exam, Part II - SOLUTIONS
- Consider the function f(x) = x
6
− 2 x
3
f(x) = x
6
− 2 x
3
f(x) = x
6
− 2 x
3
on the interval [− 2 , 2]
[− 2 , 2]
[− 2 , 2].
(a) Find the x
x x- and y
y y-coordinates of any and all local extrema and classify each as a local
maximum or local minimum.
f
′
(x) = 6x
5
− 6 x
2
0 = 6x
2
(x
3
⇒ x = 0, 1
− 2 ≤ x < 0 0 < x < 1 1 < x ≤ 2
f
′
negative negative positive
f ↘ ↘ ↗
y-values: f(0) = 0, f(1) = − 1
So, f has a local minimum at (1, −1); (0, 0) is not a local extremum.
(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.
y-values: f(−2) = 80, f(1) = −1, f(2) = 48
So, f has a global minimum at (1, −1) and a global maximum at (− 2 , 80).
(c) Find the x
x x-coordinate(s) of any and all inflection points.
f
′′
(x) = 30x
4
− 12 x
0 = 6x(5x
3
⇒ x = 0,
3
x < 0 0 < x <
3
3
- 4 < x
f
′′
positive negative positive
f concave up concave down concave up
So, the x-values of the inflection points of f are x = 0 and x =
3
- 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.
- What (if anything) does the Extreme Value Theorem say about f(x) = x
2
f(x) = x
2
f(x) = x
2
on each of the
following intervals?
EVT: If f is continuous on [a, b], then f has both a maximum and a minimum on [a, b].
(a) [1[1[1,,, 4]4]4]
f has a maximum and a minimum on [1, 4]
(b) (1(1(1,,, 4)4)4)
The EVT doesn’t apply because (1, 4) is not a closed interval since its endpoints are not included.
- Find the value of the constant c
c c that the Mean Value Theorem specifies for f(x) = x
3
f(x) = x
3
3
on [0, 3]
[0, 3]
[0, 3].
MVT: If f is continuous on [a, b] and differentiable on (a, b), then there is a number c between a and
b such that f
′
(c) =
f(b) − f(a)
b − a
For our function, we have
f(3) − f(0)
And f
′
(x) = 3x
2
′
(c) = 3c
2
So, we solve 3c
2
- (The other solution, x = −
3, is not in our interval
[0, 3].)
- 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.
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
- Find the following.
(a) all antiderivatives of 1 + 2x + x
3
x +
x
5
1 + 2x + x
3
x
x
5
1 + 2x + x
3
x +
x
5
Any such antiderivative will take the form x + x
2
x
4
x
3 / 2
x
− 4
+ C.
Note that we have used the facts that
x = x
1 / 2
and 1/x
5
= x
− 5
(b)
7
1
x
dx
7
1
x
dx
7
1
x
dx = 3 ln |x|
7
1
= 3 ln 7 − 3 ln 1 = 3 ln 7
(c)
2
− 2
4 − x
2
dx
2
− 2
4 − x
2
dx
2
− 2
4 − x
2
dx =
π(2)
2
= 2π This integral represents the area of a semicircle of radius 2.
(d)
d
dx
x
1
sin
t dt
d
dx
x
1
sin
t dt
d
dx
x
1
sin
t dt = sin
x The derivative of the area function is the original function.
