Question

If F(x)=4x-1 and g(x)=x^2+7, what is g(f(x))

Answer 1

Answer

$g(f(x)) = = 16 {x}^{2} - 8x + 8$

step by step Explanation

The functions given are

$f (x)= 4x- 1$

$g(x)= {x}^{2} + 7$

We want to find:

$g(f (x))= g( 4x- 1)$

This implies that

$g(f (x))= ( (4x- 1)^{2} + 7)$

Let us now simplify

$( {a - b)}^{2} = {a}^{2} -2ab + {b}^{2}$

This implies that

$( (4x- 1)^{2} + 7) =( {(4x)}^{2} - 2(4x) \times 1 + 1$

Combine the terms to get,

$= 16 {x}^{2} - 8x + 1 + 7$

$= 16 {x}^{2} - 8x + 8$

Answer 2

$\bf \begin{cases} f(x)=4x-1\\ g(x)=x^2+7 \end{cases}~\hspace{7em}g(~~~f(x)~~~)=[~f(x)~]^2+7 \\\\\\ g(~~~f(x)~~~)=[~4x-1~]^2+7\implies g(~~~f(x)~~~)=(4x-1)(4x-1)+7 \\\\\\ g(~~~f(x)~~~)=\stackrel{\mathbb{F ~O~ I~ L}}{(16x^2-8x+1)}+7\implies g(~~~f(x)~~~)=16x^2-8x+8$