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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 $f(f(x)) = 5$?

 Jul 24, 2020
 #1
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Im guessing AOPS?

 Jul 24, 2020
 #2
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yeah...? do you have an answer by any chance?

AnimalMaster  Jul 24, 2020
 #3
avatar+25868 
+1

x = 3 , -3               ( you would THINK     x = 2   would work, but   2 >= -4   so f(2) = 22-4=0 )

 

so f(x)  has to equal 3  or -3

 

 

x^2-4 = 3       then x  = +-sqrt7

x^2-4 = -3      then x  = +- 1           4 answers I believe........                  edited

 

 

 Jul 24, 2020
edited by ElectricPavlov  Jul 25, 2020

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