Question

How do you find `(hog)(x)` given `h(x)=x^2-2` and `g(x)=4x+1`?

Answer 1

`(h@g)(x)= 16x^2+8x-1`

Explanation:

For ease of visualisation set `g(x)=y=4x+1`

So `h(y)=y^2-2`

but `y=4x+1` so by substitution we have:

`(h@g)(x)= (4x+1)^2-2`

`(h@g)(x)= (16x^2+8x+1)-2`

`(h@g)(x)= 16x^2+8x-1`