# Algebra 2 Composition of Functions?

• Jul 24th 2012, 01:45 PM
woahitzme
Algebra 2 Composition of Functions?
If g(f(x))= 2x+1, f(x)= (1/4)x-1, then what does g(x) equal? The answer is supposed to be 8x+9. Can anyone explain to me clearly step by step how to do this? Thanks!
• Jul 24th 2012, 01:47 PM
richard1234
Re: Algebra 2 Composition of Functions?
"Fit" the expression of f(x) into g(f(x)):

$8f(x) = 8(\frac{1}{4}x - 1) = 2x - 8$

$8f(x) + 9 = (2x - 8) + 9 = 2x + 1 = g(f(x))$

Therefore, $g(x) = 8x + 9$.
• Jul 24th 2012, 02:08 PM
Soroban
Re: Algebra 2 Composition of Functions?
Hello, woahitzme!

Quote:

$\text{If }g(f(x))\:=\: 2x+1,\;\; f(x)\:=\:\tfrac{1}{4}x-1$
. . $\text{then what does }g(x)\text{ equal?}$
$\text{The answer is supposed to be: }\,8x+9.$

We are told that: . $f(x) \:=\:\tfrac{1}{4}x-1$

Then: . $g\left(\tfrac{1}{4}x-1\right) \:=\:2x+1$

Assume that $g(x)$ is a linear function: . $g(x) \:=\:ax+b$

Then: . $g\left(\tfrac{1}{4}x-1\right) \:=\:a\left(\tfrac{1}{4}x-1\right) + b \;=\;2x + 1$

We have: . $\tfrac{a}{4}x - a + b \:=\:2x+1 \quad\Rightarrow\quad \tfrac{a}{4}x + (b-a) \:=\:2x+1$

Equate coefficients: . $\begin{array}{ccc}\dfrac{a}{4} &=& 2 \\ b-a &=& 1 \end{array}$

Hence: . $a \,=\,8,\;b\,=\,9$

Therefore: . $g(x) \:=\:8x + 9$