+0  
 
-1
68
2
avatar+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
avatar+74 
-1

I believe it's 5.

 

cool

 Feb 21, 2021
 #2
avatar+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

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