# For the function g(x) = 32/(x+3), what is (g^-1 cdot g)(5)?

Jul 1, 2017

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{68}{5}$

#### Explanation:

Let's start by finding an equation for the inverse.

$y = \frac{32}{x + 3}$

Switch the x and the y.

$x = \frac{32}{y + 3}$

$x \left(y + 3\right) = 32$

$x y + 3 x = 32$

$x y = 32 - 3 x$

$y = \frac{32 - 3 x}{x}$

Now let's complete the required operation.

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{32 - 3 x}{x} \cdot \frac{32}{x + 3}$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{32 - 15}{5} \cdot \frac{32}{5 + 3}$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{17}{5} \cdot 4$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{68}{5}$

Hopefully this helps!

Jul 1, 2017

$\left({g}^{- 1} \cdot g\right) \left(5\right) = 5$

#### Explanation:

It should be obvious that irrespective of the definition of the function $g$ then the composition:

$\left({g}^{- 1} \cdot g\right) \left(x\right) \equiv x$

That is, in words, if we apply any function to a variable $x$, then apply the inverse function the we end up with what we started with, i.e. $x$

If you are not convinced, then we can easily demonstrate this by determining an expression for the inverse function ${g}^{- 1} \left(x\right)$:

$g \left(x\right) = \frac{32}{x + 3} \implies g = \frac{32}{x + 3}$
$\therefore x + 3 = \frac{32}{g}$
$\therefore x = \frac{32}{g} - 3$

So for the function $g \left(x\right) = \frac{32}{x + 3}$, the inverse function ${g}^{- 1} \left(x\right)$ is defined by:

${g}^{- 1} \left(x\right) = \frac{32}{x} - 3$

Now let us find $g \left(5\right)$

$g \left(5\right) = \frac{32}{5 + 3} = \frac{32}{8}$

Now let us find the composition $\left({g}^{- 1} \cdot g\right) \left(5\right)$

$\left({g}^{- 1} \cdot g\right) \left(5\right) = {g}^{- 1} \left(g \left(5\right)\right)$
$\text{ } = {g}^{- 1} \left(\frac{32}{8}\right)$
$\text{ } = \frac{32}{\frac{32}{8}} - 3$
$\text{ } = 8 - 3$
$\text{ } = 5$