How do you find the compositions given #f(x) = 2x + 3# and #h(x) = 2x^2 + 2x + 1#?
1 Answer
Explanation:
Let's start with the composition
To calculate this, you basically need to insert
#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,
Similarly, here, you need to plug
#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
# = 2(4x^2 + 2 * 2x * 3 + 3^2) + 4x + 6 + 1#
# = 8x^2 + 24x + 18 + 4x + 6 + 1#
# = 8x^2 + 28x + 25#