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If $f(x)=\large\frac{x}{x+1},$ $g(x)=\large\frac{x}{1-x}$ and $h(x)=\sqrt [3] {x},$ then $(fogoh)^{-1}(8)= ?$

(A) 1

(B) 2

(C) 8

(D) 512

1 Answer

$goh(x)=\large\frac{\sqrt [3] {x}}{1-\sqrt [3] {x}}$
$fogoh(x)=\large\frac{\large\frac{\sqrt [3] {x}}{1-\sqrt [3] {x}}}{\large\frac{\sqrt [3] {x}}{1-\sqrt [3] {x}}+1}$= $\sqrt [3] {x}$
Let $fogoh(x)=y$
$\Rightarrow x=(fogoh)^{-1}y$
$or\:\:\sqrt [3] {x}=y$
$\Rightarrow\:x=y^3$
$\Rightarrow\:(fogoh)^{-1}(8)=8^3=512$
answered May 21, 2013 by rvidyagovindarajan_1
 

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