Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 31
(b) ln x +ln(x − 1) = ln(x(x − 1)) = 1 ⇔ x(x − 1) = e 1 ⇔ x 2 − x − e =0. The quadratic formula (with a =1,
b = −1,andc = −e)givesx = 1 2
1 ±
√ 1+4e
, but we reject the negative root since the natural logarithm is not
defined for xe −1 ⇒ x>e −1 ⇒ x ∈ (1/e, ∞)
53. (a) For f(x) = √ 3 − e 2x ,wemusthave3 − e 2x ≥ 0 ⇒ e 2x ≤ 3 ⇒ 2x ≤ ln 3 ⇒ x ≤ 1 2
ln 3. Thus, the domain
of f is (−∞, 1 2
ln 3].
(b) y = f(x) = √ 3 − e 2x [note that y ≥ 0] ⇒ y 2 =3− e 2x ⇒ e 2x =3− y 2 ⇒ 2x =ln(3− y 2 ) ⇒
x = 1 2 ln(3 − y2 ). Interchange x and y: y = 1 2 ln(3 − x2 ).Sof −1 (x) = 1 2 ln(3 − x2 ). For the domain of f −1 ,wemust
have 3 − x 2 > 0 ⇒ x 2 < 3 ⇒ |x| < √ 3 ⇒ − √ 3