Question

How do you find the compositions given `f(x) = 2x + 3` and `h(x) = 2x^2 + 2x + 1`?

Answer 1

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

`h(f(x)) = 8x^2 + 28x + 25`

Explanation:

Let's start with the composition `f(h(x))`.

To calculate this, you basically need to insert `h(x)` for every occurence of `x` in `f(x)`:

`f(color(violet)(x)) = 2color(violet)(x) + 3`

`f(color(blue)(h(x))) = 2color(blue)(h(x)) + 3 = 2(color(blue)(2x^2 + 2x + 1)) = 4x^2 + 4x + 2`

Now, let's do the other composition, `h(f(x))`.

Similarly, here, you need to plug `f(x)` for every occurence of `x` in `h(x)`:

`h(color(violet)(x)) = 2color(violet)(x)^2 + 2color(violet)(x) + 1`

`h(color(blue)(f(x))) = 2[color(blue)(f(x))]^2 + 2color(blue)(f(x)) + 1`

`= 2(color(blue)(2x+3))^2 + 2(color(blue)(2x+3)) + 1`

`= 2(2x+3)^2 + 2(2x+3) + 1`

... I'm using the formula `(a+b)^2 = a^2 + 2ab + b^2`...

`= 2(4x^2 + 2 * 2x * 3 + 3^2) + 4x + 6 + 1`

`= 8x^2 + 24x + 18 + 4x + 6 + 1`

`= 8x^2 + 28x + 25`