NAME:
MATH 151
November 5, 2014
QUIZ 7
• Calculators are NOT allowed!
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Find (f −1 )0 (2) where
f (x) = e2x + 4x + 1.
Solution:Use the formula (f −1 )0 (a) =
1
f 0 (f −1 (a))
(f −1 )0 (2) =
where a = 2 in this case. So
1
f 0 (f −1 (2))
.
Since you cannot explicitly compute f −1 , to compute f −1 (2), we find an x such that f (x) = 2.
Using x = 0 gives f (0) = e0 + 0 + 1 = 2 so f −1 (2) = 0.
The derivative of f is f 0 (x) = 2e2x + 4. Putting all of this together gives
(f −1 )0 (2) =
1
f 0 (f −1 (2))
=
1
1
1
=
=
.
f 0 (0)
2e0 + 4
6
2. (3 points) Solve for x:
log10 (5 − x) + log10 (2 − x) = 1
Solution:Using properties of logarithms,
1 = log10 (5 − x) + log10 (2 − x)
= log10 ((5 − x)(2 − x))
= log10 (10 − 7x + x2 )
so 1 = log10 (10 − 7x + x2 ). Taking 10 raised to each side gives:
2
101 = 10log10 (10−7x+x ) = 10 − 7x + x2
and so 10 = 10 − 7x + x2 ⇒ x2 − 7x = 0 ⇒ x(x − 7) = 0. Our candidates for x are x = 0
and x = 7. Plugging x = 7 in the original equation gives log10 (−2) and log10 (−5), both of
which are undefined. Thus x = 0 .
NAME:
MATH 151
3. (3 points) Compute the limit:
lim ln(3x2 + 5) − 2 ln(x + 1)
x→∞
Solution: Using properties of logarithms:
ln(3x2 + 5) − 2 ln(x + 1) = ln(3x2 + 5) − ln((x + 1)2 )
2
3x + 5
= ln
(x + 1)2
3x2 + 5
.
= ln
x2 + 2x + 1
Thus
3x2 + 5
2
lim ln(3x + 5) − 2 ln(x + 1) = lim ln
x→∞
x→∞
x2 + 2x + 1
3x2 + 5
= ln lim 2
x→∞ x + 2x + 1
= ln(3)
November 5, 2014