College Algebra, 2013a

452 Exponential and Logarithmic Functions
(+) 0 (−)
−1
0 (+)
4
y = r(x) =2 x2 −3x − 16
e
2. The first step we need to take to solve x
e x −4
≤ 3 is to get 0 on one side of the inequality. To
that end, we subtract 3 from both sides and get a common denominator
e x
e x − 4
≤ 3
e x
e x − 4 − 3 ≤ 0
e x
e x − 4 − 3(ex − 4)
e x − 4
≤ 0 Common denomintors.
12 − 2e x
e x − 4
≤ 0
We set r(x) = 12−2ex
e x −4
and we note that r is undefined when its denominator e x − 4 = 0, or
when e x = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To
find the zeros of r, wesolver(x) = 0 and obtain 12 − 2e x = 0. Solving for e x , we find e x =6,
or x = ln(6). When we build our sign diagram, finding test values may be a little tricky since
we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing 4
which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the
sign of r (ln(3)) may be very unsettling, remember that e ln(3) =3,so
r (ln(3)) =
12 − 2eln(3)
e ln(3) − 4
=
12 − 2(3)
3 − 4
= −6
We determine the signs of r (ln(5)) and r (ln(7)) similarly. 5 From the sign diagram, we
find our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of
f(x) =
ex
e x −4
is below the graph of g(x) =3on(−∞, ln(4)) ∪ (ln(6), ∞), and they intersect
at x =ln(6)≈ 1.792.
4 This is because the base of ln(x) ise>1. If the base b were in the interval 0