SOLUTIONS TO EXAM 1, MATH 10560
1. Suppose f (x) is a one-to-one function on its domain and f (2) = 8, f (8) = 3, f 0 (2) = 4,
and f 0 (8) = 7. Find (f −1 )0 (8).
Answer : 14
Solution: Since (f −1 )0 (f (x)) = f 01(x) , 8 = f (2) and f 0 (2) = 4
(f −1 )0 (8) = (f −1 )0 (f (2)) =
1
1
= .
f 0 (2)
4
2. Solve the following equation for x:
ln 5 − e2x = 1 .
√
Answer : x = ln 5 − e
Solution:
ln(5 − e2x ) = 1 ⇐⇒ 5 − e2x = e ⇐⇒ e2x = 5 − e
√
1
⇐⇒ 2x = ln(5 − e) ⇐⇒ x = ln(5 − e) ⇐⇒ x = ln 5 − e.
2
3. Find the equation of the tangent line to f (x) at x = 1 if
√
f (x) = e
x+6
.
e7
Answer : y = e7 + 2 (x − 1)
Solution: Notice that the slope of this tangent line is f 0 (x) and for x = 1 we have
√
1
1 0
x+6 1
f (x) x=1 = e
= e7 .
· ·√ 2
x x=1 2
The equation of the tangent line is
1
y = f (1) + f 0 (x)(x − 1) ⇐⇒ y = e7 + e7 (x − 1).
2
4. Evaluate the following definite integral.
Z e 3
2x + x + 1
dx .
x2
1
Answer : 1 + e2 − e−1
Solution:
Z
1
e
2x3 + x + 1
dx =
x2
=
e
Z
2x +
1
e
x2 1
2
1
1
+ 2 dx
x x
e
+ ln x|e1 + (−x−1 )1
= e + 1 − e−1 .
1
2
SOLUTIONS TO EXAM 1
5. A sample of radioactive material decays to 1/20 of its original amount in one day.
How long would it take to decay to 1/100 of its original amount?
Answer : ln(100)
ln(20)
1
Solution: Let t be the time (measured in days) needed to decay to 100
. Then
(
1 t
1
− ln(100)
ln(100)
) =
⇐⇒ t =
=
.
20
100
− ln(20)
ln(20)
6. Calculate the following definite integral.
Z 1
7x
dx .
x
0 7 +5
ln 2
Answer : ln
7
Solution: Using the u-substitution u = 7x + 5, so that du = ln 7 · 7x dx and then
Z 12
Z 1
1 1
1
7x
dx =
· du =
ln u|12
6
x
ln 7 u
ln 7
6
0 7 +5
1
12
ln 2
1
(ln 12 − ln 6) =
(ln ) =
.
=
ln 7
ln 7
6
ln 7
7. Find f 0 (x) for
f (x) = ln(x4 + 2) + arcsin(x−1 ) .
3
Answer : x4x
4 +2 −
Solution:
√ 1
x4 −x2
f 0 (x) =
=
4x3
1
−1
+p
· 2
2
x +2
1 − (x−1 )2 x
1
4x3
−√
.
+2
x4 − x2
x2
8. Calculate the following integral.
Z
arctan x
dx .
1 + x2
2
x)
+C
Answer : (arctan
2
dx
Solution: Use the u-substitution u = arctan x, then du = 1+x
2 and therefore
Z
Z
arctan x
1
dx = udu = u2 + C
2
1+x
2
(arctan x)2
=
+C
2
9. Evaluate
Answer :
Z
1
5
sec5 x − 13 sec3 x + C
tan3 (x) sec3 (x)dx .
SOLUTIONS TO EXAM 1
3
Solution: Set u = sec x, then du = tan x · sec xdx. By u-substitution we have:
Z
Z
3
3
tan (x) sec (x)dx = tan2 x · sec x2 · (tan x sec x)dx
Z
1
1
= (1 − u2 )u2 du = u3 − u5 + C
3
5
1
1
= sec5 x − sec3 x + C.
5
3
2
Z
x3 ln x dx.
10. Evaluate
1
Answer: 4 ln 2 − 15
16
Solution: Set u = ln x and dv = x3 dx, then du = x1 dx and v = 14 x4 . Using integration
by parts we have:
Z 2
Z 2
1 4 2
1 4 1
3
x ln x dx = ln x · x −
x · dx
4 1
x
1
1 4
Z 2
1
= 4 ln 2 −
x3 dx
4 1
15
= 4 ln 2 − .
16
11. Use logarithmic differentiation to find f 0 (x) if
√
x cos3 (x + 7)
f (x) =
.
(3x2 + 2)6
Solution: Since
ln f (x) =
1
ln x + 3 ln cos(x + 7) − 6 ln(3x2 + 2),
2
f 0 (x) = f (x) · (ln f (x))0 and
(ln f (x))0 =
we have
√
f 0 (x) =
1
3
6
+
· − sin(x + 7) − 3
· 6x,
2x cos(x + 7)
3x + 2
x cos3 (x + 7)
1
36x
·(
+ 3 tan(x + 7) − 3
)
(3x2 + 2)6
2x
3x + 2
12. Evaluate the limit
1
lim (cos x) x2 .
x→0
Solution: Note that
1
lim (cos x)
x→0
1
x2
lim ln((cos x) x2 )
= ex→0
.
4
SOLUTIONS TO EXAM 1
Since
1
lim ln((cos x) x2 ) = lim
x→0
x→0
ln cos x
0
=
x2
0
− sin x
cos x
− tan x
0
=
x→0 2x
x→0
2x
0
− sec2 x
1
(L’Hospital) = lim
=− ,
x→0
2
2
(L’Hospital) = lim
we have
= lim
1
lim (cos x)
x→0
1
x2
lim ln((cos x) x2 )
= ex→0
1
= e− 2 .
13. Evaluate:
x2
dx.
4 − x2
Solution: We use the trigonometric substitution x = 2 sin θ, then dx = 2 cos θdθ so
that
Z
Z
(2 sin θ)2
x2
√
p
dx =
· 2 cos θdθ
4 − x2
4 − (2 sin θ)2
Z
Z
2
= 4 sin θdθ = 2(1 − cos 2θ)dθ
Z
√
= 2θ − sin 2θ + C
x
= 2 arcsin( ) − 2 sin θ cos θ + C
2
x
1p
= 2 arcsin( ) − x ·
4 − x2 + C.
2
2