# Question #33262

Oct 30, 2016

#### Explanation:

Given:

$f \left(x\right) = \frac{2}{x + 1}$

And:

$g \left(f \left(x\right)\right) = \frac{8}{x + 1}$

Find g(x):

Begin by finding ${f}^{-} 1 \left(x\right)$, because we know that:

$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = g \left(x\right)$

Find ${f}^{-} 1 \left(x\right)$:

Substitute ${f}^{-} 1 \left(x\right)$ for $x$ in $f \left(x\right)$:

$f \left({f}^{-} 1 \left(x\right)\right) = \frac{2}{{f}^{-} 1 \left(x\right) + 1}$

By definition, substitute x for $f \left({f}^{-} 1 \left(x\right)\right)$:

$x = \frac{2}{{f}^{-} 1 \left(x\right) + 1}$

Multiply both sides of the equation by $\frac{{f}^{-} 1 \left(x\right) + 1}{x}$

${f}^{-} 1 \left(x\right) + 1 = \frac{2}{x}$

Subtract 1 from both sides:

${f}^{-} 1 \left(x\right) = \frac{2}{x} - 1$

Substitute $\frac{2}{x} - 1$ into $g \left(f \left(x\right)\right)$:

$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\left(\frac{2}{x} - 1\right) + 1}$

Remove the ()s in the denominator:

$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\frac{2}{x} - 1 + 1}$

-1 + 1 is zero:

$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = \frac{8}{\frac{2}{x}}$

Perform the division:

$g \left(f \left({f}^{-} 1 \left(x\right)\right)\right) = 8 \left(\frac{x}{2}\right)$

Write the left side as $g \left(x\right)$ and simplify the right:

$g \left(x\right) = 4 x$