+0

# Piecewise-Defined Functions

-1
68
2
+204

If

$f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases}$

then for how many values of $x$ is $f(f(x)) = 5$?

Feb 21, 2021

#1
+74
-1

I believe it's 5.

Feb 21, 2021
#2
+112966
+2

$$f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases}$$

then for how many values of    $$x$$   is  $$f(f(x)) = 5$$  ?

---------

You need to look at 2 different cases.

first assume f(x) is less than -4

solve

5=x+3

x=-2

-2 is not less than -4 so no solutions here.

now assume f(x) >= -4

$$5=x^2-4\\ 9=x^2\\ x=\pm3$$

Both of these are bigger than -4 so they both could have answers.  That is, for this problem

f(x) could be +3 or -3

Now you need to look at each of these for each of the scenarios  x<-4 and x>=-4  and see how many answers you get that are valid.

Feb 22, 2021