- b.(ex+ e–x)^2 = (ex)^2 + 2(ex)(e–x) + (e–x)^2 = e^2 x
+ 2 + e–2x= e^2 x+ e–2x+ 2
- d. = = = 5^8 x–4=
(5^4 )^2 x–1 = 625^2 x–1
- b.ex(ex–1) – e–x(ex– 1) = (ex)^2 – ex– (e–x)(ex) –
(e–x)(–1) = e^2 x– ex– 1+ e–x
- b.
=
=
=
= e^4 x+ 1
- b.Ifb1, the graph ofy= bxgets very close
to the x-axis as the x-values move to the left. A
typical graph is as follows:
762. d.If 0b1, the graph ofy= bxgets very
close to the x-axis as the x-values move to the
right, and the y-values grow very rapidly as the
x-values move to the left. A typical graph is as
follows:
763. c.Ifb0, then bx0, so the equation bx= 0
has no solution.
764. c.Note that ^12
- x
= 2x. The graph ofy= 2xis
always above the x-axis, so statement c is true.
- b.Observe that 1 – 3x 0 is equivalent to
3 x 1. The graph ofy= 3xis always increasing
and is equal to 1 when x= 0, so for all xthere-
after, 3x 1.Hence, the solution set is [0,∞).
- c.Observe that –^23
2 x
= ^49
x
, which is strictly
positive, for any real number x. Therefore, the
solution set for the inequality –^23
2 x
0 is the
empty set.
–10 –8 –6 –4 –2 2 4 6 8 10
–2
–4
–6
–8
–10
10
8
6
4
2
–10 –8 –6 –4 –2 2 4 6 8 10
–2
–4
–6
–8
–10
10
8
6
4
2
e^2 x+ e–2x
e–2x
e^2 x– 1 + 1 + e–2x
e–2x
(ex)(ex) – (ex)(e–x) + (e–x)(ex) + (e–x)(e–x)
e–2x
ex(ex– e–x) + e–x(ex+ e–x)
e–2x
^510 x–4
52 x
^59 x–3^5 x–1
52 x
(5^3 x–1)^3 ^5 x–1
52 x
ANSWERS & EXPLANATIONS–
