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# Piecewise-Defined Function

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

Feb 4, 2021
#3
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Are you sure? How did you arrive at this answer?

Guest Feb 4, 2021
#2
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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