# If f(x)=2x-7 and fog(x)=4x+3 find (gof)^(-1)(x)?

Mar 7, 2018

${\left(g \circ f\right)}^{- 1} \left(x\right) = \frac{x + 9}{4}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 2 x - 7$
and since
$\textcolor{w h i t e}{\text{XXX}} f \circ g \left(x\right) = f \left(g \left(x\right)\right) = 2 \cdot g \left(x\right) - 7$

But we are also told that
$\textcolor{w h i t e}{\text{XXX}} f \circ g \left(x\right) = 4 x + 3$

So
$\textcolor{w h i t e}{\text{XXX}} 2 \cdot g \left(x\right) - 7 = 4 x + 3$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 2 \cdot g \left(x\right) = 4 x + 10$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow g \left(x\right) = 2 x + 5$

Therefore
color(white)("XXX")gcircf(x)=(g(f(x))
$\textcolor{w h i t e}{\text{XXXXXXxX}} = 2 \cdot f \left(x\right) + 5$
$\textcolor{w h i t e}{\text{XXXXXXxX}} = 2 \left(2 x - 7\right) + 5$
$\textcolor{w h i t e}{\text{XXXXXXxX}} = 4 x - 14 + 5$
$\textcolor{w h i t e}{\text{XXXXXXxX}} = 4 x - 9$

For simplicity, let $y = g \circ f \left(x\right)$
and we are asked to find ${\left(g \circ f\right)}^{- 1} \left(x\right) = {y}^{- 1}$

Given an equation with $y$ defined in terms of $x$
the easiest way to find the inverse, ${y}^{- 1}$ is to solve the equations
to define $x$ in terms of $y$
then replace $x$ with ${y}^{- 1}$ and $y$ with $x$:

We have
$\textcolor{w h i t e}{\text{XXX}} y = 4 x - 9$

$\textcolor{w h i t e}{\text{XXX}} 4 x = y + 9$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{y + 9}{4}$

then doing the replacement
$\textcolor{w h i t e}{\text{XXX}} {y}^{- 1} = \frac{x + 9}{4}$

that is
$\textcolor{w h i t e}{\text{XXX}} {\left(g \circ f\right)}^{- 1} \left(x\right) = \frac{x + 9}{4}$