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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?

 Aug 24, 2020
 #1
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Could i please have a hint, rather than the whole answer?

 Aug 24, 2020
 #2
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First think of the inner f(x) as, say, y.

 

f(y) = 5    

so either y2 - 4 = 5 or y + 3 = 5  

If y2 - 4 = 5, then y = 3 is a possibility as 3 > -4, also y = -3 is a possibility as -3 > -4

If y + 3 = 5 then  y = 2, which is not possible as 2 is not less than -4

 

So there are two possibilities for y, which means you have to consider two possibilities for f(x)

 

(1)  f(x) = 3

      3 = x2 - 4  means x2 = 7  or x = ±√7  both of which are > -4

       3 = x + 3  means x = 0  which is not possible because 0 is not less than -4.

 

(2) f(x) = -3

     Can you do this one?

 Aug 24, 2020

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