# Decomposing A Composite Function

Discussion in 'Other Pre-University Math' started by nycmathguy, Aug 3, 2021.

1. ### nycmathguy

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Section 1.8
Question 50

See attachment.

I know that (f ° g)(x) = f(g(x)).

We have f(g(x)) = h(x).

This leads to f(g(x)) = (1 - x)^3.

Stuck here....

nycmathguy, Aug 3, 2021
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2. ### MathLover1

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This leads to f(g(x)) = (1 - x)^3 => g(x)=1 - x and f(x)=x^3

MathLover1, Aug 3, 2021
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3. ### nycmathguy

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How did you conclude that f(x) = x^3?

nycmathguy, Aug 4, 2021
4. ### MathLover1

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f(g(x)) -> f after g(x)
f(g(x)) = (1 - x)^3=> on the left side, g(x) is inside parentheses and on the right side 1 - x is inside parentheses , means
g(x)=1 - x which means (1 - x) represents x in f(x)
then f(x )= x^3

MathLover1, Aug 4, 2021
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5. ### nycmathguy

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This is not clear to me. Sorry. I will try two more on my own. If I get it right, I may try a few more samples. If not, I am moving on.

nycmathguy, Aug 4, 2021
6. ### MathLover1

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"Function Composition" is applying one function to the results of another:
-----> f( ) --------> g( ) ------>
The result of f() is sent through g()
It is written: (g º f)(x)
Which means: g(f(x))

Example: f(x) = 2x+3 and g(x) = x^2

"x" is just a placeholder. To avoid confusion let's just call it "input":

f(input) = 2(input)+3

g(input) = (input)^2

(g º f)(x) = g(f(x))

First we apply f, then apply g to that result:

-----> (2*input+3 ) --------> (input )^2 ------>

(g º f)(x) = (2x+3)^2

your case is in reverse order

f(g(x)) = (1 - x)^3 =>f(input) = (input)^3 =>"x" is= "input" =>f(x) = (x)^3

MathLover1, Aug 5, 2021
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