# Given that f(x) = x2 − 4x − 3 and g(x) = the quantity of x plus three, over four , solve for f(g(x)) when x = 9? −6 3 6 9

## Given that f(x) = x2 − 4x − 3 and $g \left(x\right) = \frac{x + 3}{4}$, solve for $f \left(g \left(x\right)\right)$ when $x = 9$? 6 ; 3; 6; 9

Jun 22, 2018

$- 6$

#### Explanation:

$\text{to evaluate "f(g(x))" substitute "x=g(x)" into } f \left(x\right)$

$g \left(x\right) = \frac{x + 3}{4}$

$f \left(g \left(x\right)\right) = f \left(\frac{x + 3}{4}\right)$

$= {\left(\frac{x + 3}{4}\right)}^{2} - \cancel{4} \left(\frac{x + 3}{\cancel{4}}\right) - 3$

$= \frac{{\left(x + 3\right)}^{2}}{16} - \left(x + 3\right) - 3$

$f \left(g \left(\textcolor{red}{9}\right)\right) = \frac{{\left(\textcolor{red}{9} + 3\right)}^{2}}{16} - \left(\textcolor{red}{9} + 3\right) - 3$

$\textcolor{w h i t e}{\times \times x} = \frac{144}{16} - 12 - 3$

$\textcolor{w h i t e}{\times \times x} = 9 - 15 = - 6$

Jun 22, 2018

I will only do the first one for you. It has a full explanation

$f \left(g \left(9\right)\right) = - 6$

#### Explanation:

Given that $g \left(\textcolor{m a \ge n t a}{x}\right) = \frac{\textcolor{m a \ge n t a}{x} + 3}{4} \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} g \left(\textcolor{m a \ge n t a}{9}\right) = \frac{\textcolor{m a \ge n t a}{9} + 3}{4}$

So wherever there is an $x$ put a $\textcolor{m a \ge n t a}{9}$ instead.

$g \left(9\right) = \frac{12}{4} = \textcolor{g r e e n}{3}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that $f \left(x\right) = {x}^{2} - 4 x - 3$

So wherever there is a $g \left(x\right)$ put a $\textcolor{g r e e n}{3}$

$f \left(g \left(x\right)\right) = f \left(\textcolor{w h i t e}{\text{d")ubrace(g(color(magenta)(9)))color(white)("d}}\right)$
$\textcolor{w h i t e}{\text{dddddddddddd}} \downarrow$
$f \left(g \left(x\right)\right) = f \left(\textcolor{w h i t e}{\text{d.d")color(green)(3)color(white)("dd.}}\right)$

$\textcolor{b r o w n}{f \left(\textcolor{g r e e n}{3}\right) = {\textcolor{g r e e n}{3}}^{2} \textcolor{w h i t e}{\text{dd}} \underbrace{- 4 \left(\textcolor{g r e e n}{3}\right) - 3}}$
color(brown)(color(white)("ddddddddddd.d")darr
$f \left(3\right) = + 9 \textcolor{w h i t e}{\text{d")-15color(white)("ddd}} = - 6$