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Unformatted text preview: Solutions to Quiz 3A
www.math.uﬂ.edu/˜harringt
January 24, 2008
1. Find the exact value of each expression:
(a) log5 125 = log5 53 = 3 log5 5 = 3
(b) ln(1/e) = ln(1) − ln(e) = 0 − 1 = −1
√
2. Find the inverse of f (x) = 3 − e2x and ﬁnd the range of f −1 .
(a) First we will ﬁnd the inverse function.
f (x) = y
x
x2
x2 − 3
3 − x2
ln(3 − x2 ) =
=
=
=
=
= √
3 − e2x
√
3 − e2y
3 − e2y
−e2y
e2y
2y 1
ln(3 − x2 ) = y
2
1
ln(3 − x2 )
f −1 (x) =
2
(b) Next we will ﬁnd the range of f −1 by ﬁnding the domain of f .
Recall that the range of f −1 is the same as the domain of f .
Since f (x) = √ 3 − e2x then we need to ﬁnd when 3 − e2x ≥ 0
3 − e2x
3
ln 3
1
ln 3
2
1 ≥0
≥ e2x
≥ 2x
≥x So the range of f −1 is (−∞, 1 ln 3].
2
3. The point P (1, 2) lies on the curve y = x2 + 1. If Q is the point
(x, x2 + 1), then ﬁnd the slope of the secant line for:
(a) x = 0
Find the slope between P (1, 2) and Q(0, 1). So:
m= 2−1
=1
1−0 (b) x = 2
Find the slope between P (1, 2) and Q(2, 5). So:
m= 2 2−5
=3
1−2 ...
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This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Math, Calculus, Geometry

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