Question

Given `g(x) = sqrt(5x - 4)` and `h(x) = 4x^2 + 7` how do you find h(g(1))?

Answer 1

You must plug 1 in for x inside its respective function. Once you get the result from that calculation you must plug it into the adjacent function.

Explanation:

`h(g(1)) = 4(sqrt(5(1) - 4))^2 + 7`

`h(g(1)) = 4(1)^2 + 7`

`h(g(1)) = 4 + 7`

`h(g(1)) = 11`

Don't forget to work from inside to the outside. Example: `f(g(h(x))) =`h inside g inside f

Practice exercises:

1. Evaluate the following compositions if `f(x) = 2^(x/2), g(x) = 3x^2 - 4x + 1 and h(x)=sqrt(2x + 9`

a). `h(g(f(x)))`

b). `g(f(g(h(x))))`

c). `g(h(f(6)))`

Good luck!

Answer 2

`h(g(1))=11`

Explanation:

Doing one step at a time!

`color(blue)("Consider g(1)")`

`g(x)=sqrt(5x-4)`

`color(blue)(color(magenta)(g(1))= sqrt(5(1)-4) = sqrt(5-4)=sqrt(1)=color(red)(+-1))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider "h(g(1))`

`h(x)=4x^2+7`

`h(g(x))=4(g(x))^2+7`

`h(g(1))=4(color(magenta)(g(1)))^2+7`

`h(g(1))=4(color(red)(+-1))^2+7`

But `(-1)^2=(+1)^2 =color(green)( +1)`

`h(g(1))=4(color(green)(+1))^2+7`

`h(g(1))=4+7 =11`