Given $f (x) = \frac{1}{x - 2}$ $f(x)=1/(x-2)$ , how do you find f(f(x))?

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Answer 1 · 2016-03-31T15:13:40

$f (f (x)) = \frac{x - 2}{2 x + 5}$ $f(f(x))=" "(x-2)/(2x+5)$

Where ever you see $x$ $x$ replace it with $\frac{1}{x - 2}$ $1/(x-2)$

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

$Consider just the denominator$ $color(blue)("Consider just the denominator")$

Multiply the 2 by $1$ $1$ but in the form of $1 - \frac{x - 2}{x - 2}$ $1-(x-2)/(x-2)$

So the denominator becomes

$\frac{1}{x - 2} - (\frac{2}{1} \times \frac{x - 2}{x - 2})$ $1/(x-2) -(2/1xx(x-2)/(x-2))$

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

$\frac{1 + 2 x + 4}{x - 2}$ $(1+2x+4)/(x-2)$

$\frac{2 x + 5}{x - 2}$ $(2x+5)/(x-2)$

$Substitute into original expression$ $color(blue)("Substitute into original expression")$

$f (f (x)) = 1 \div \frac{2 x + 5}{x - 2} = \frac{x - 2}{2 x + 5}$ $color(green)(f(f(x))= 1 -: (2x+5)/(x-2)" "=" "(x-2)/(2x+5))$

Given f(x)=1x−2f(x)=1/(x-2), how do you find f(f(x))?