SOLUTION: Given the function y = 2cos(3x) + 1, determine the three functions f(x), g(x), and h (x) so that f°g°h(x) = 2cos(3x) + 1. b) Consider the functions f(x) = 2x

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Question 1194808: Given the function y = 2cos(3x) + 1, determine the three functions f(x), g(x), and h (x) so that f°g°h(x) = 2cos(3x) + 1. b) Consider the functions f(x) = 2x — 1 and g(x) = x2. For what value(s) of x does f°g(x) =g°f(x)?

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52794) (Show Source):
You can put this solution on YOUR website!
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Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

a) This question is not posed correctly. There is no answer as to what are "THE three functions" f(x), g(x), and h(x) for which f(g(h(x)))=2cos(3x)+1. There are many combinations of three functions for which the composition of the functions gives that result.

Consider what you would do to evaluate the function 2cos(3x)+1 for a given value of x. The stages of the computation would be
x --> 3x --> cos(3x) --> 2cos(3x) --> 2cos(3x)+1

The operations performed on the input variable are
(1) multiply by 3
(2) evaluate the cosine
(3) multiply by 2
(4) add 1

You could write the given function as a composition of four functions by having each of those functions perform one of those four operations:
k(x) = 3x h(x) = cos(x) --> h(k(x))=cos(3x) g(x) = 2x --> g(h(k(x)))=2cos(3x) f(x) = x+1 --> f(g(h(k(x))))=2cos(3x)+1

There are many different ways you could get the given function using the composition of three functions by combining any two of the four operations in one function. For example,
h(x)=cos(3x) g(x)=2x --> g(h(x))=2cos(3x) f(x)=x+1 --> f(g(h(x)))=2cos(3x)+1

But there are many other ways to form the given function as a composition of three functions. An example could be
h(x)=2cos(3x) g(x)=x --> g(h(x))=2cos(3x) (the function g does nothing....) f(x)=x+1 --> f(g(h(x)))=2cos(3x)+1

Since there are many ways to form the given function as a composition of three functions, it is impossible to know what the expected answer is to this question.

b) NOTE: Use "^" (shift-6) to represent exponentiation in typed text: g(x)=x^2, not just "x2"

f(x)=2x-1
g(x)=x^2

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

If f(g(x))=g(f(x)), then

ANSWER: f(g(x))=g(f(x)) for x=1 only

SOLUTION: Given the function y = 2cos(3x) + 1, determine the three functions f(x), g(x), and h (x) so that f°g°h(x) = 2cos(3x) + 1. b) Consider the functions f(x) = 2x — 1 and g(x) = x2.