Math 105a
Final Exam
Solutions
1. (a) If f (x) =
sin x
then
x
f 0 (x) =
g 0 (t) =
(b) If g(t) = ln(sin(t2 )) then
Z
(c)
Z
t3 + 7t
dt =
t4
π/4
(d)
0
Z x→0
1
+ 7t−3
t
cos(t2 ) · 2t
= 2t cot(t2 )
sin(t2 )
dt = ln |t| −
7
+C
2t2
√
π/4
2
cos x dx = sin x
= sin (π/4) − sin 0 =
2
0
cos x − 1
ex − x − 1
(e) lim
x cos x − sin x
x2
L0 H
=
lim
x→0
− sin x − 1
ex − 1
L0 H
=
lim
x→0
− cos x
= −1
ex
2. Use the limit definition of the derivative to find f 0 (x) for f (x) =
f 0 (x)
f (x + h) − f (x)
= lim
h→0
h→0
h
= lim
1
2+x+h
−
h
1
2+x
1
.
2+x
(2 + x) − (2 + x + h)
h→0 h(2 + x)(2 + x + h)
= lim
−h
−1
1
= lim
=−
h→0 h(2 + x)(2 + x + h)
h→0 (2 + x)(2 + x + h)
(2 + x)2
= lim
(
− cos x,
3. Consider f (x) =
x2 − a,
if x < 0
if x ≥ 0
(a) lim− f (x) = lim− (− cos x) = −1
x→0
x→0
(b) lim f (x) = lim (x2 − a) = −a
x→0+
x→0+
(c) In order for f (x) to be continuous at x = 0 we require lim− f (x) = lim+ f (x) = f (0),
x→0
x→0
note: f (0) = −a.
lim f (x) = lim+ f (x)
x→0−
⇐⇒
−1 = −a
⇐⇒
a=1
x→0
(d) Let f be defined with a = 1. Use a graphical approach to zoom in on the graph of f near x = 0.
From (c) we know the function is continuous. Close inspection of the graph reveals no sharp
corners or vertical tangents at x = 0. More specifically, we can see that as x approaches zero from
both the left and right that the slope of the function is zero. Therefore, f 0 (0) = 0.
Note: this can be shown more formally using the limit definition of derivative.
1
4. A 216 m2 rectangular pea-patch is to be enclosed by a fence and divided into two equal parts by
another fence parallel to one of the sides. What dimensions for the outer rectangle will require the
smallest total length of fence? How much fencing material will be needed?
We want to minimize the total length of fence.
Let x represent the width of the rectangle. Let y
represent the length of the rectangle. We need to
minimize P = 3x + 2y. The area of the rectangle
given by A = xy is fixed at 216 m2 . Substitute
216
for y into the formula for perimeter
y =
x 216
P = 3x + 2
.
x
3x2 − 432
dP
432
= 0 when x = ±12.
=3− 2 =
dx
x
x2
Given the context of the problem, x = 12 is the only appropriate critical value.
Therefore,
d2 P
864
=
> 0, therefore the total length of fence is minimized when x = 12 meters.
dx2
(12)3
The dimensions of the outer rectangle are x = 12 meters and y = 18 meters.
The amount of fence material is P = 3(12) + 2(18) = 72 meters.
At x = 12,
5. The tangent line to the curve x2/3 +y 2/3 = 8 at the point (8, 8) and the coordinate axes form a triangle,
as shown in the picture below. What is the area of the triangle?
Find the equation of the tangent line in order to
locate the x and y-intercepts to help determine
the base and height of the triangle. To find the
slope of the tangent line, implicitly differentiate
dy
both sides with respect to x, solve for
, and
dx
then substitute in the x and y-values at (8, 8) .
r
x
dy
x1/3
=⇒
= − 1/3 = − 3
dx
y
y
r
dy
3 8
The slope of the tangent line at (8, 8) is given by
=−
= −1.
dx
8
2 −1/3 2 −1/3 dy
x
+ y
·
=0
3
3
dx
The equation of the tangent line is y = −x + 16. Therefore, the x and y-intercepts are (16, 0) and
1
(0, 16), respectively. The area of the triangle is A = (16)(16) = 128.
2
6. Recall the Mean Value Theorem: if f is continuous on [a, b] and differentiable on (a, b), then there
f (b) − f (a)
exists a number c, with a < c < b, such that f 0 (c) =
.
b−a
The function f (x) = x2 + 2x − 1 satisfies the conditions of the Mean Value Theorem on [0, 1] because
f is continuous on [0, 1] and f 0 (x) = 2x + 2 exists on (0, 1). Therefore, it follows that there exists a
f (1) − f (0)
number c, with 0 < c < 1, such that f 0 (c) =
= 3.
1−0
1
f 0 (c) = 2c + 2 = 3 =⇒ c = .
2
2
Z
7. (15 pts) Let F (x) =
x
√
ln t dt for x ≥ 2.
2
(a) Apply the Fundamental Theorem of Calculus Part II:
Z x
√
√
d
F 0 (x) =
ln t dt = ln x.
dx
2
√
3
Z
(b) To approximate F (3) =
ln t dt using a left sum with n = 3 we pick ∆t =
2
1
.
3
√
Since the graph of f (t) = ln t is increasing then the left sum will provide a lower estimate. A
table of values may be created to help compute the left and right sums.
t
f (t)
2
0.833
7/3
0.920
8/3
0.990
3
1.048
1
(0.833 + 0.920 + 0.990) ≈ 0.9143
3
√
(c) F increasing for x ≥ 2 because F 0 (x) = ln x > 0 for x ≥ 2.
Left Sum =
(d) F 00 (x) > 0 because F 0 (x) is an increasing function. Therefore, F is concave up.
8. The graph of f 0 (x) is given below. Goal: to construct the antiderivative f with f (0) = 2.
Z
(a) To find f (1), let a = 0 and b = 1. Applying the FTC gives:
1
f 0 (x)dx = f (1) − f (0)
0
1
Z
f 0 (x)dx = 2 + 1 = 3
Therefore, f (1) = f (0) +
0
2
Z
f 0 (x)dx = 3 + 0.5 = 3.5
Similarly, f (2) = f (1) +
1
3
Z
f 0 (x)dx = 3.5 + (−0.5) = 3
f (3) = f (2) +
2
3.5
Z
f 0 (x)dx = 3 + (−0.25) = 2.75
f (3.5) = f (3) +
3
Z
4
f 0 (x)dx = 3 + 0 = 3
f (4) = f (3) +
3
Z
5
f (5) = f (4) +
f 0 (x)dx = 3 + 1.5 = 4.5
4
x
f (x)
0
2
1
3
2
3.5
3
3
3
3.5
2.75
4
3
5
4.5
(not required, but useful in (b))
(b) Inflection points occur when the function changes concavity.
The graph of f 0 (x) has sharp corners at x = 1, x = 3, and x = 4, so f 00 (x) does not exist at those
values. Therefore, f has possible inflection points at x = 1, x = 3, and x = 4.
We can use a sign chart to see that f 00 (x) < 0 for 1 < x < 3 since f 0 (x) is decreasing on that
interval. Similarly, f 00 (x) > 0 for 3 < x < 4 and 4 < x < 5 since f 0 (x) is increasing on each
interval.
Therefore, the graph of f is concave down on (1, 3), and concave up on (3, 4) and (4, 5).
(c) The graph of f (x)
•
•
•
•
is linear for 0 < x < 1 since the graph of f 0 is constant on that interval
has a local maximum at (2, 3.5) since f 0 changes from positive to negative at x = 2
has a local minimum at (3.5, 2.75) since f 0 changes from negative to positive at x = 3.5
changes from concave down to concave up at x = 3
The graph is provided below.
9. Suppose the rate of growth of a tumor is given by
Z
(a) The integral
dr
= 0.3et mm/month.
dt
3
0.3et dt represents the total amount, in mm, that the tumor has grown through-
0
out the first three months.
Z
1 3
0.3et dt represents the average rate of growth, in mm/month, of the tumor for
3 0
the first three months.
(b) The integral
(c) The general solution is r(t) = 0.3et + C.
(d) Use r(0) = 5 to find C:
5 = 0.3e0 + C
=⇒
C = 4.7
t
The particular solution is r(t) = 0.3e + 4.7.
(e) Evaluate r(3) to find the size of the tumor after 3 months
r(3) = 0.3e3 + 4.7 ≈ 10.73 mm.
OR use part (a) along with the initial value in (c): the total size of the tumor is equal to the
initial size at t = 0 plus the amount of growth in the first three months
Z 3
3 5+
0.3et dt = 5 + 0.3et = 5 + 0.3e3 − 0.3 ≈ 10.73 mm.
0
0
4