+193 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 $f(f(x)) = 3$?
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. | ||||
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Through all this algebra, we have determined that there are 3 values of x for which \(f(f(x)) = 3\).
