How do you find (f of g of h) if #f(x)=x^2+1# #g(x)=2x# and #h(x)=x-1#?

Question

How do you find  (f of g of h)  if #f(x)=x^2+1# #g(x)=2x# and #h(x)=x-1#?

Caleb Alexander · Accepted Answer

undefined To find $ (f \circ g \circ h)(x) $ where $ f(x) = x^2 + 1 $, $ g(x) = 2x $, and $ h(x) = x - 1 $, you need to perform the composition in the following order: $ f(g(h(x))) $.

1. First, find $ h(x) $ applied to $ x $: $ h(x) = x - 1 $.
2. Then, find $ g(x) $ applied to the result of step 1: $ g(h(x)) = g(x - 1) = 2(x - 1) = 2x - 2 $.
3. Finally, find $ f(x) $ applied to the result of step 2: $ f(g(h(x))) = f(2x - 2) = (2x - 2)^2 + 1 = 4x^2 - 8x + 5 $.

So, $ (f \circ g \circ h)(x) = 4x^2 - 8x + 5 $.

Liam Brennan · Answer

#(f@g@h)(x) = 4x^2-8x+5#

Given:

#{ (f(x) = x^2+1), (g(x) = 2x), (h(x) = x-1) :}#

One way of thinking about these function compositions is to go back and forth between the symbols and verbal descriptions of what the functions do.

In our example:

#f# takes the square of a number and adds #1#

#g# doubles a number

#h# subtracts #1# from a number

So a verbal description of the composed #f@g@h# as a sequence of steps might be:

Subtract #1#

Double

Square

Add #1#

So in symbols we might describe this process thus:

#x -> x-1 -> 2(x-1) -> (2(x-1))^2 -> (2(x-1))^2+1#

So:

#(f@g@h)(x) = f(g(h(x)))#

#color(white)((f@g@h)(x)) = (2(x-1))^2+1#

#color(white)((f@g@h)(x)) = 4(x^2-2x+1)+1#

#color(white)((f@g@h)(x)) = 4x^2-8x+4+1#

#color(white)((f@g@h)(x)) = 4x^2-8x+5#