# If f(x)=4[x-1] and g(x)=x+1, how do you find f(g(-1))?

Sep 11, 2015

$f \left(g \left(- 1\right)\right) = - 4$

#### Explanation:

When you have an expression:
$f \left(y\right)$ ,
this means that you should plug in whatever $y$ is into $x$ in the function $f \left(x\right)$. That is, replace every instance of $x$ with $y$.

There are two ways to find $f \left(g \left(- 1\right)\right)$
The first is to solve from the inside out.

Plug in $- 1$ into $g \left(x\right)$ to get $g \left(- 1\right)$:
$g \left(x\right) = x + 1$
$g \left(- 1\right) = - 1 + 1$
$\textcolor{b l u e}{g \left(- 1\right) = 0}$

Then, plug in $g \left(- 1\right)$ into $f \left(x\right)$:
$f \left(x\right) = 4 \left[x - 1\right]$
$f \left(g \left(- 1\right)\right) = 4 \left[0 - 1\right]$
$\textcolor{b l u e}{f \left(g \left(- 1\right)\right) = - 4}$

The second way is to get $f \left(g \left(x\right)\right)$ first, then plug in $- 1$:

$\textcolor{red}{f \left(x\right) = 4 \left[x - 1\right]}$
$\textcolor{b l u e}{g \left(x\right) = x + 1}$

$\textcolor{red}{f \left(\textcolor{b l u e}{g \left(x\right)}\right)} = \textcolor{red}{4 \left[\textcolor{b l u e}{x + 1} - 1\right]}$

Then, you can plug in $- 1$

$f \left(g \left(- 1\right)\right) = 4 \left[- 1 + 1 - 1\right]$
$\textcolor{b l u e}{f \left(g \left(- 1\right)\right) = - 4}$