Question

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

Answer 1

`\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`

Answer 2

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) * 1 + 1`

Combine the terms to get,


`= 16 {x}^(2) - 8x + 1 + 7`


`= 16 {x}^(2) - 8x + 8`