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+10
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avatar+1132 

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 \(f(f(x)) = 3\)?

 Jul 12, 2022
 #1
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0

I graphed it!  There are 5 solutions.  LOL

 Jul 12, 2022
 #2
avatar+1132 
+7

unfortunately that's wrong.

nerdiest  Jul 12, 2022
 #3
avatar+252 
+1

I think it would be a good idea if you solve this for yourself. :)

 

Do some casework!

Set up some cases such as x >= 3, and where x < 3. Remember, just because x >= 3, doesn't mean that f(x) is also greater than or equal to 3, set up some cases there too! There may also be extraneous solutions! Get rid of those :)

 Jul 12, 2022

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