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If 

\[f(x) =
\begin{cases}
2x-5 &\quad \text{if } x \ge 3, \\
-x + 5 &\quad \text{if } x < 3,
\end{cases}
\]

then for how many values of $x$ is $f(f(x)) = 3$?

 Feb 4, 2021
 #1
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0

The equation f(f(x)) = 3 has 1 solution.

 Feb 4, 2021
 #3
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0

Are you sure? How did you arrive at this answer?

Guest Feb 4, 2021
 #2
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+1

This problem seems overwhelming, but let's take this problem one piece at a time.

 

Instead of trying to wrap our heads around \(f(f(x))=3\), we should instead ponder a far easier question. For what values of x does \(f(x) = 3\)? Well, let's investigate, shall we?

 

\(f(x)\) is defined as a piecewise function, so we must more or less treat it as two separate functions and ensure that our answers lie within the restrictions.

 

\(2x-5=3\\ 2x=8\\ x_1=4\\ 4\geq 3 \checkmark\) \(-x+5=3\\ -x=-2\\ x_2=2\\ 2<3\checkmark\) I placed checkmarks around both proposed solutions because they respect the boundaries given by the original piecewise function.

 

\(f(x_1)=3\text{ and } f(x_2)=3\). Let and solve\(f(x) = x_1\text{ or } f(x) = x_2\)to determine the number of solutions to \(f(f(x)) = 3\).

 

\(f(f(x))=3\\ f(x_1)=3\\ f(x)=x_1\\ f(x)=4\) \(f(f(x))=3\\ f(x_2)=3\\ f(x)=x_2\\ f(x)=2\) Let's find all the values for x that satisfy these two equations. This will require 4 equations in total.
\(2x-5=4\\ 2x=9\\ x_{11}=\frac{9}{2}\\ \frac{9}{2}\geq3\checkmark\) \(-x+5=4\\ -x=-1\\ x_{12}=1\\ 1<3\checkmark\)

 

\(2x-5=2\\ 2x=7\\ x=\frac{7}{2}\\ \frac{7}{2}\geq3\checkmark\) \(-x+5=2\\ -x=-3\\ x=3\\ 3\nless 3\\ \therefore \text{Invalid solution}\)

 

 

 

Through all this algebra, we have determined that there are 3 values of x for which \(f(f(x)) = 3\).

 Feb 4, 2021

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