Name:
TA’s Name/Section #:
Recitation Day/Time:
Math 165: Sample Final
Part I
Spring 2013
This part of the exam has 10 problems; each problem is worth 10 points.
You may NOT use a calculator on this section. You must show all work, but you need
not simplify your answers. This part of the exam will be collected after 70 minutes.
Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Question 8:
Question 9:
Question 10:
100 Total Points:
Question 1 (10 points). Given f (x) = ex (x2 − x − 11):
(a) Find all values of x such that f 0 (x) = 0.
(b) Find all critical points.
(c) Find the intervals where f is increasing and where f is decreasing.
(d) Find all local maxima and local minima.
(e) Find all inflection points.
Solution.
(a) x = −4; and x = 3
9
(b) −4, 4 or about (−4, 0.1648); and (3, −5e3 ) or about (3, −100.4277)
e
(c) Increasing: (−∞, −4) ∪ (3, ∞) Decreasing: (−4, 3)
9
(d) −4, 4 is a local maximum and (3, −5e3 ) is a local minimum.
e
√
3
53
(e) Inflection points at x = − ±
2
2
Page 2
Question 2 (10 points). A moving car applies its brakes, producing a constant deceleration
of 40 ft/s2 . It produces skid marks measuring 160 feet before coming to a stop. How fast was
the car traveling when the brakes were applied?
Solution. Starting from a(t) = −40 and taking two antiderivatives, we arrive at the velocity
and position functions:
v(t) = −40t + v0
s(t) = −20t2 + v0 t + s0
We can choose our initial position to be the point at which the brakes are applied, so that
s0 = 0 and these becomes:
v(t) = −40t + v0
s(t) = −20t2 + v0 t
We also know that the time at which the velocity is zero is the same time that the position
is equal to 160. So setting v(t) = 0 and s(t) = 160 gives us two equations, which we can
solve for either t or v0 :
0 = −40t + v0
160 = −20t2 + v0 t
160 = −20t2 + (40t)t
160 = 20t2
8 = t2
√
t=2 2
√
0 = −40(2 2) + v0
√
v0 = 80 2 ft/s
Page 3
Question 3 (10 points). Find the derivative of:
f (x) = ln(3x − 2)3
Solution.
f 0 (x) =
3(3x − 2)2 (3)
9(3x − 2)2
9
=
=
(3x − 2)3
(3x − 2)3
3x − 2
Question 4 (10 points). Find the antiderivative:
Z 2
x +x
dx
2x − 1
Solution. Since the numerator has higher degree than the denominator, we can use longdivision to rewrite the integrand and solve:
Z 2
Z
x +x
1
3
3
dx =
x+ +
dx
2x − 1
2
4 4(2x − 1)
1
3
3
= x2 + x + ln(2x − 1) + C
4
4
8
Page 4
Question 5 (10 points). Find the integral:
Z
e−1/x
dx
x2
Solution.
u=−
du =
Z
1
x
1
dx
x2
eu du = eu + C
= e−1/x + C
Z
Question 6 (10 points). Evaluate
7
√
x + 2 dx.
2
Solution.
38
3
Page 5
Question 7 (10 points). Find the integral:
Z
1/2
0
Solution.
sin−1 x
√
dx
1 − x2
π2
72
Question 8 (10 points). Find the integral:
Z 3
|3x − 5| dx
0
Solution.
41
6
Page 6
Question
points). Find the average value of the function f (x) = sin2 x cos x on the
i
h 9π (10
π
interval − , .
2 4
Solution.
1
+
π
4
Z
π
2
√
2+4
sin2 x cos x dx =
9π
Z
20π
| sin x| dx.
Question 10 (10 points). Use periodicity to evaluate
0
Solution.
| sin(x + π)| = | − sin x| = | sin x|
Therefore,
Z
20π
Z
π
| sin x| dx
| sin x| dx = 20
0
0
Z
= 20
sin x dx
0
= 40
Page 7
π
Name:
TA’s Name:
Recitation Day/Time:
Math 165: Sample Final
Part II
Spring 2013
This part of the exam has 8 problems; each problem is worth 10 points.
Answer each question completely. Show all work. No credit is allowed for mere answers
with no work shown. Show the steps of calculations. State the reasons that justify any
conclusions you make.
Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Question 8:
70 Total Points:
Page 8
Question 1 (10 points). A forest ranger is in a forest 2 miles from a straight road. The
rangers car is located five miles down the road. If the forest ranger can walk 3 miles per hour
in the forest and 4 miles per hour along the road, towards what point on the road should the
ranger walk to minimize the time needed to walk to the car?
6√
7 miles, or approximately 2.3 miles, down the road.
7
Solution. The distance is
Question 2 (10 points). Solve the differential equation and find that solution for which
y = 2 when x = 0:
x3
dy
=
dx
1−y
Solution. Separate the variables and integrate
Z
Z
(1 − y)dy =
x3 dx
y−
Determine C : 2 −
22
2
The solution is: y −
x4
y2
=
+C
2
4
= 0 + C so that C = 0.
y2
2
=
x4
.
4
Page 9
Find the area of the region under the curve y = f (x) over the given interval. To do this,
divide the interval into n equal subintervals, calculate the Riemann sum corresponding to
such a partition and then let n → ∞. You may need to use the following formula:
n
X
i=1
i2 =
n(n + 1)(2n + 1)
6
Question 3 (10 points).
1
f (x) = x2 + 2 on [0, 1]
2
Solution. Fix ∆x = n1 , xi =
i
n
Rp =
so that the corresponding Riemann sum is:
n
X
i=1
n
X
1 i2
1
f (xi )∆x =
( 2 + 2)
2n
n
i=1
n
n
n
X
1 i2
2
1 1 X 2 2X
=
( 3 + )=
1
i +
2n
n
2 n3 i=1
n i=1
i=1
1 1 n(n + 1)(2n + 1) 2
+ n
2 n3
6
n
(n + 1)(2n + 1)
=
+2
12n2
=
Finally: limn→∞ Rp =
2
12
+2=
13
.
6
Z
Question 4 (10 points). Find the interval(s) on which the graph y = f (x) =
x ≥ 0, is (a) increasing, and (b) concave up.
Solution.
(a) f is increasing on [0, 2] and [4, ∞)
(b) f is concave up on (3, ∞).
Page 10
0
x
(t2 − 6t + 8) dt,
1
Question 5 (10 points). Find the area bounded by the curve y = x − 1, the lines x = 4
2
and x = 8, and the x-axis. Do this in two ways:
(a) using plane geometry,
(b) using an antiderivative and the Fundamental Theorem of Calculus.
Solution. Both methods give an area of 8.
Question 6 (10 points). Find the integral:
Z
tan(2x) + x dx
Solution.
− ln | cos 2x| + x2
+C
2
Page 11
Question 7 (10 points). Find f −1 , and check your answer by finding f −1 (f (x)), if:
f (x) = −
10
x + 12
Solution. Set y = f (x), and then switch x and y to obtain:
x=−
10
y + 12
Then solve for y:
xy + 12x = −10
xy = −10 − 12x
10 + 12x
y=−
x
We can now compose the two:
10
10 + 12 − x+12
f −1 (f (x)) = −
10
− x+12
10(x + 12) − 120
=
10
10x
=
10
=x
Page 12
Question 8 (10 points). Polonium-214 has a very short half-life of 1.4 × 10−4 seconds. If a
sample has an initial mass of 50 mg, how many grams remain after a hundredth of a second?
Solution. Even if we did not know a formula for half-life, we do know that it is based on
exponential decay, and that after 1.4 × 10−4 seconds, we should have a mass of 25 mg. We
can use this information to find the parameter k:
m(t) = 50 · ekt
25 = 50 · ek(0.00014)
1
= e(0.00014)k
2
1
ln = (0.00014)k
2
− ln 2
=k
0.00014
−4951.05 ≈ k
Now we input this k and the evaluate our formula at t = 10−2 :
m(0.01) ≈ 50 · e(−4951.05)(0.01)
≈ 50 · 3.14676 × 10−22
≈ 1.57338 × 10−20
Page 13