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Math 105: Review for Final Exam, Part II - SOLUTIONS
- Consider the function f(x) = x
3
ln x
f(x) = x
3
ln x f(x) = x
3
ln x on the interval [1/e, e
2
]
[1/e, e
2
]
[1/e, e
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) = 3x
2
ln x + x
3
x
0 = x
2
(3 ln x + 1)
⇒ x
2
= 0 (never on 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 xxx- and yyy-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 (never on 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
- You are watching a plane flying toward your position at a constant height of 3 miles and
a speed of 500 miles per hour relative to the ground. At the moment when the plane is
5 miles from you (diagonally), at what rate is the angle of your vision toward the plane
changing?
a
c
θ
Plane
You
We know
da
dt
, and we want to find
dθ
dt
So, we write an equation that relates a and θ and then differentiate implicitly with respect to time t.
tan θ =
a
sec
2
θ
dθ
dt
a
2
da
dt
dθ
dt
a
2
da
dt
cos
2
θ
At the moment in question, c = 5, so by the Pythagorean Theorem we know that a = 4 and that
cos θ =
. Finally, we are told that
da
dt
So,
dθ
dt
2
2
= 60 radians per hour.
- 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.
- 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, 4]
[1, 4]
[1, 4]
f has a maximum and a minimum on [1, 4]
(b) (1, 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[0[0,,, 3]3]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
- Find the following.
(a)
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
(b)
4
1
(1 + 2x + x
3
x +
x
5
) dx
4
1
(1 + 2x + x
3
x +
x
5
) dx
4
1
(1 + 2x + x
3
x +
x
5
) dx =
[
x + x
2
x
4
x
3 / 2
x
− 4
]
4
1
2
4
3 / 2
− 4
2
4
3 / 2
− 4
Note that we have used the facts that
x = x
1 / 2
and 1/x
5
= x
− 5
(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.
(e) lim
n→∞
n
n
k=
2 k
n
2
lim
n→∞
n
n
k=
2 k
n
2
lim
n→∞
n
n
k=
2 k
n
2
This represents the limit of a right-hand sum as the number (n) of rectangles goes to infinity.
As k goes from 1 to n, the expression
2 k
n
takes on the values
n
n
2 n
n
; the first of these values is just to the left of 1 and the last is equal to 3, so we see that
we are looking at the function on the interval [1, 3].
The
n
out front is our ∆x, which confirms that we are dealing with an interval of length 2 being
subdivided into n equal subintervals.
Finally, each of the x-values is being squared, so the function in question must be f(x) = x
2
Thus, we see that the expression is the area under f(x) = x
2
on [1, 3].
Its value is
3
1
x
2
dx =
x
3
3
1
3
3
- Water is leaking out of a tank at a decreasing rate r(t)
r(t) r(t) as shown below.
time (min) 000 222 444 666 888
rate (gal/min) 15
(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 net change (in gallons) of the amount of water in the tank on the interval
[2, 6] minutes.
- Consider the graph of f(t)
f(t) f(t) 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 G(2)
G(2)
G(2), G(4)
G(4)
G(4), and H(4)
H(4)
H(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 GGG increasing? Where is GGG 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).
