Question #33262

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Answer 1 · 2016-10-30T16:46:59

Please see the explanation.

Given:

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

And:

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

Find g(x):

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

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

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

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

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

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

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

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

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

Subtract 1 from both sides:

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

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

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

Remove the ()s in the denominator:

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

-1 + 1 is zero:

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

Perform the division:

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

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

$g (x) = 4 x$ $g(x) = 4x$