Question

What is `h(f(-2))` given `f(x)=x^2` and `h(x)=x+4`?

Answer 1

`h(f(-2)) = 8`

Explanation:

To solve `color(blue)(h(color(red)(f(-2))))` we can first solve the function `color(red)(f(-2))`

`color(red)(f(-2)) = (-2)^2 = 4`

`color(blue)(h(color(red)(f(-2))))` now becomes `color(blue)(h(color(red)(4)))`

`color(blue)(h(color(red)(4))) = 4 + 4 = 8`