APPM 1345
Exam 3 Solutions
Spring 2015
1. (8 points) Match the graphs shown to four of the following functions.
No explanation is necessary.
(a) y = −e
(b) y = e
x
�
� �
(d) y = log4 x
−x
−1
(e) y = sin
�
x
(f) y = tan−1 x
(c) y = ln x
Solution: (1) e (2) f (3) c (4) d
2. (32 points) Evaluate the following.
√ √
3
(a) Find f 0 23 if f (u) = ln
cos−1 u . Simplify your answer.
(b) If g(x) = Cax , g(1) = 3, and g(3) = 34 , find the positive constants C and a.
√
(c) Find dy/dx given x2y = ( y)x , x > 0, y > 0. Leave your answer unsimplified.
2
(d) Find the local maximimum and minimum values of h(t) = ln(t + 5) , if any.
Solution:
(a)
f (u) = ln
f0
√
1
f 0 (u) = ·
3
√ !
3
1
= ·
2
3
1
ln cos−1 u
3 1
−1
√
cos−1 u
1 − u2
!
1
−1
4
p
= −
π/6
π
1/4
3
cos−1 u
=
(b) We are given
4
g(1) = Ca = 3, g(3) = Ca3 = .
3
Divide the two equations.
4
2
⇒ a=
9
3
2
9
g(1) = Ca = C
=3 ⇒ C=
3
2
a2 =
�
�
(c) Use logarithmic and implicit differentiation.
√
x2y = ( y)x = y x/2
x
2y ln x = ln y
2
1
dy
x 1 dy 1
2y · + 2 ln x
= ·
+ ln y
x
dx
2 y dx 2
dy
x dy
1
2y
2 ln x
−
= ln y −
dx 2y dx
2
x
dy
x
1
2y
2 ln x −
= ln y −
dx
2y
2
x
1
ln y −
dy
= 2
dx
2 ln x −
2y
x
x
2y
(d)
h(t) = ln(t + 5)
2
⇒ h0 (t) = 2 ln(t + 5) ·
2 ln(t + 5)
1
=
t+5
t+5
Solve for the critical numbers where h0 = 0. The derivative h0 is defined for all t in (−5, ∞), the
domain of h.
2 ln(t + 5)
= 0 ⇒ ln(t + 5) = 0 ⇒ t = −4
t+5
Use the first derivative test to determine whether
there is a maximum or minimum value at the
critical number t = −4. Since h0 (− 29 ) = 4 ln 12 < 0 and h0 (−3) = ln 2 > 0, there is an absolute
minimum value of h(−4) = (ln 1)2 = 0 and no maximum values.
h0 (t) =
3. (20 points)
Z e6
dx
(a)
x log6 x
e
sec2 e−3x
dx
e3x
Z
(b)
Z
(c)
tan−1 (2t)
dt
1 + 4t2
Solution:
(a) First convert to natural log. Then let u = ln x, du = dx/x. The new limits are then u = 1 to 6.
Z
e
e6
dx
=
x log6 x
Z
e
e6
ln 6
dx = (ln 6)
x ln x
Z
6
1
i6
du
= (ln 6) ln |u| = (ln 6)(ln 6 − ln 1) = (ln 6)2
u
1
(b) Let u = e−3x , du = −3e−3x dx.
Z
Z
sec2 e−3x
1
1
1
dx = −
sec2 u du = − tan u + C = − tan e−3x + C
3x
e
3
3
3
(c) Let u = tan−1 (2t), du = 2 dt/ 1 + 4t2 .
Z
tan−1 (2t)
1
dt =
2
1 + 4t
2
Z
u du =
2
1
1 u2
·
+C =
tan−1 (2t) + C
2 2
4
4. (20 points) Let f (x) =
ln x
.
2 + 3 ln x
(a) Find the domain of f .
(b) Evaluate lim f (x).
x→0+
(c) Find an equation for the line tangent to y = f (x) at x = 1e .
(d) Show that f is one-to-one.
(e) Find the inverse function f −1 .
Solution:
(a) The function is undefined if x = e−2/3 or x ≤ 0. The domain of f is (0, e−2/3 ) ∪ (e−2/3 , ∞).
(b) lim
x→0+
ln x
= lim
2 + 3 ln x x→0+
2
ln x
2
1
= 1/3 since lim ln x = −∞ and lim
= 0.
x→0+
x→0+ ln x
+3
(c)
(2 + 3 ln x) x1 − ln x ·
f 0 (x) =
(2 + 3 ln x)2
2
1
= 1
= 2e
f0
2
e
e (2 + 3(−1))
1
−1
f
=
=1
e
2 + 3(−1)
The tangent line is y = 1 + 2e x −
1
e
3
x
=
2 + 3 ln x − 3 ln x
2
=
2
x(2 + 3 ln x)
x(2 + 3 ln x)2
= 2ex − 1.
(d) Because x > 0 and (2 + 3 ln x)2 > 0, the derivative f 0 is always positive. Therefore f is an
increasing function and one-to-one.
(e) Solve for x, then swap x and y.
ln x
2 + 3 ln x
2y + 3y ln x = ln x
y=
2y = (1 − 3y) ln x
2y
ln x =
1 − 3y
x = e2y/(1−3y)
f −1 (x) = e2x/(1−3x)
5. (20 points) Use these approximations to calculate your answers to the following problems:
ln 2 ≈ 0.7, ln 3 ≈ 1.1, ln 5 ≈ 1.6, ln 11 ≈ 2.4.
(a) How long will it take an investment of D dollars to quadruple in value if the interest rate is 5% per
year compounded continuously?
(b) Superman is locked in a room with a kryptonite-like substance, which renders him powerless. His
strength will be restored once the radioactive substance has disintegrated by two-thirds. This will
take 22 hours. What is the half-life of the substance?
Solution:
(a) Let A(t) equal the amount of the investment after t years. Find t when A(t) = 4D.
A(t) = De0.05t = 4D
e0.05t = 4
0.05t = ln 4
2 ln 2
2(0.7)
140
ln 4
=
≈
=
= 28 years
t=
0.05
0.05
0.05
5
(b) Let m(t) equal the amount of kryptonite after t hours. We are given that m(22) = 31 m(0). First
find the rate constant k.
m(t) = m(0)ekt
1
m(22) = m(0)e22k = m(0)
3
1
e22k =
3
22k = ln(1/3)
ln 3
k=−
22
Now find t when m(t) = 12 m(0).
1
m(t) = m(0)ekt = m(0)
2
1
ekt =
2
kt = ln(1/2)
ln 2
−22
t=−
= − ln 2
k
ln 3
22
7(22)
≈ (0.7) ·
=
= 14 hours
1.1
11