Question 1194808
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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.<br>
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<br><pre>
x --> 3x --> cos(3x) --> 2cos(3x) --> 2cos(3x)+1</pre>
The operations performed on the input variable are
(1) multiply by 3
(2) evaluate the cosine
(3) multiply by 2
(4) add 1<br>
You could write the given function as a composition of four functions by having each of those functions perform one of those four operations:<br><pre>
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</pre>
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,<br><pre>
h(x)=cos(3x)
g(x)=2x  -->  g(h(x))=2cos(3x)
f(x)=x+1  -->  f(g(h(x)))=2cos(3x)+1</pre>
But there are many other ways to form the given function as a composition of three functions.  An example could be<br><pre>
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</pre><br>
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.<br>
b) NOTE: Use "^" (shift-6) to represent exponentiation in typed text: g(x)=x^2, not just "x2"<br>
f(x)=2x-1
g(x)=x^2<br>
f(g(x))=2(x^2)-1 = 2x^1-1
g(f(x))=(2x-1)^2 = 4x^2-4x+1<br>
If f(g(x))=g(f(x)), then<br>
{{{2x^2-1=4x^2-4x+1}}}
{{{2x^2-4x+2=0}}}
{{{x^2-2x+1=0}}}
{{{(x-1)^2=0}}}
{{{x=1}}}<br>
ANSWER: f(g(x))=g(f(x)) for x=1 only<br>