If #f ( x ) = 2x - 9 and g ( x ) = ( x + 1) ^ { 2},# what is the composite function #f(g(x))#?

1 Answer
Dec 10, 2017

Substitute the functions into one another and get #f(g(x)) = 2(x + 1)^2 - 9 = 2x^2 + 4x - 7#.

Explanation:

We have #f(x) = 2x - 9# and #g(x) = (x+1)^2# and are asked to find #f(g(x)#. Start by putting #g(x)# into that:

#f(g(x)) = f((x+1)^2)#

Now, if #f(x) = 2x - 9#, then anything can fit inside #x# and be put into the equation. Let's set variable #k = (x+1)^2# and fit it inside #f(x)#:

#f((x+1)^2)= f(k) = 2k - 9#

Now we just replace #k# with #(x+1)^2#:

#f(k) = 2k - 9 = 2(x+1)^2 - 9#

Since essentially #k = (x+1)^2 = g(x)#, we now have #f(k) = f(g(x)) = 2(x+1)^2 - 9#.

If asked to expand, do so:

#f(g(x)) = 2(x+1)^2 - 9 = 2(x + 1)(x + 1) - 9#

#= 2(x^2 + x + x + 1) - 9 = 2(x^2 + 2x + 1) - 9#

#= 2x^2 + 4x + 2 - 9 = 2x^2 + 4x - 7#

So we have:

#f(g(x)) = 2(x + 1)^2 - 9 = 2x^2 + 4x - 7#