The inverse function of f(x) = (82x-8-2x/82x + 8-2x), xϵ (-1, 1), is .

Question

The inverse function of f(x) = (82x-8-2x/82x + 8-2x), xϵ (-1, 1), is . (A) (1/4)(log8 e) loge ((1-x/1+x)) (B) (1/4)(log8 e) loge ((1+x/1-x)) (C) (1/4) loge ((1+x/1-x)) (D) (1/4) loge ((1- x/1+ x))

Tardigrade Admin · Accepted Answer

f (x)=y=(84x-1/84x+1)=1-(2/84x+1) so, 84x+1=(2/1-y) ⇒ 84x=(1+y/1-y) ⇒ x=ℓ n((1+y/1-y))× (1/4ℓ n8)=f -1(y) Hence, f -1(x)=(1/4)logg eℓ n((1+x/1-x))

Q. The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}, x ϵ (- 1, 1),$ is __________.

$\frac{1}{4} (l o g_{8} e) l o g_{e} (\frac{1 - x}{1 + x})$

$\frac{1}{4} (l o g_{8} e) l o g_{e} (\frac{1 + x}{1 - x})$

$\frac{1}{4} l o g_{e} (\frac{1 + x}{1 - x})$

$\frac{1}{4} l o g_{e} (\frac{1 - x}{1 + x})$

Q. The inverse function of f(x)=82x+8−2x82x−8−2x​,xϵ(−1,1), is __________.

41​(log8​e)loge​(1+x1−x​)

41​(log8​e)loge​(1−x1+x​)

41​loge​(1−x1+x​)

41​loge​(1+x1−x​)

f(x)=y=84x+184x−1​=1−84x+12​ so, 84x+1=1−y2​⇒84x=1−y1+y​ ⇒x=ℓn(1−y1+y​)×4ℓn81​=f−1(y) Hence, f−1(x)=41​logg​eℓn(1−x1+x​)

Q. The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}, x ϵ (- 1, 1),$ is __________.

$\frac{1}{4} (l o g_{8} e) l o g_{e} (\frac{1 - x}{1 + x})$

$\frac{1}{4} (l o g_{8} e) l o g_{e} (\frac{1 + x}{1 - x})$

$\frac{1}{4} l o g_{e} (\frac{1 + x}{1 - x})$

$\frac{1}{4} l o g_{e} (\frac{1 - x}{1 + x})$