# Question #402e2

Jan 25, 2018

$f \circ g \left(2\right) = 9$

#### Explanation:

$f \circ g$ is read "f of g of x". It's a "composition function". All you do is put the second function in the first. So substitute $g \left(x\right)$ for the $x$ in $f \left(x\right)$, and you've got your new function. Here are two methods:
1. Find $g \left(2\right)$ first, then plug that in to $f \left(x\right)$:
$g \left(2\right) = 2 \left(2\right) + 1 = 5$
Plug that into $f \left(x\right) : f \left(5\right) = 5 + 4 = 9$.
or
2. Find $f \circ g \left(x\right)$ then plug $2$ in for $x$:
$f \circ g \left(x\right) = f \left(g \left(x\right)\right) = f \left(2 x + 1\right)$
Plugging $\left(2 x + 1\right)$ in for $x$ gives us
$f \left(2 x + 1\right) = \left(2 x + 1\right) + 4 = 2 x + 5$
and
If $f \circ g \left(x\right) = 2 x + 5$, then $f \circ g \left(2\right) = 2 \left(2\right) + 5 = 9$

The second way looks longer, but they're both handy to know. Hope this helps, and have fun!