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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 \(f(f(x)) = 5\)

 Jun 28, 2018
 #1
avatar+98129 
+3

 

I'm not that good at these, but I will give this one a try

 

Note that, for the given constraints, as small as the first function can be for any x  is  -4

So.....any value we choose for x  will be evaluated as being ≥ -4

 

So...in the first case....we are looking for  this

 

(x^2 - 4)^2  - 4  = 5

x^4 - 8x^2 + 16  - 4  = 5

x^4  - 8x^2 + 7  = 0

Factoring, we have

(x^2 - 1 ) ( x^2 - 7)  = 0

Setting both factors to 0 and solving for x  results in the values    1, -1 ,  √7 , - √7 

 

All of these make f (f(x))  = 5

 

In the other case, we need to choose an x < -4  which we can put into  x + 3  which will result in 3  or -3....note that when we put  -3 or 3 into the first function we get  5

0 will make x + 3  = 3  but x has to be < -4 to use this function

But  -  6  will work  because   f(-6)  =  (-6) + 3  =  -3

And this value when plugged into the first function to give a result of 5

That is  f ( f(-6) )  =  5

 

So...the values for x  that  will work are   1, -1 ,  √7 , - √7   and  - 6

 

 

cool cool cool

 Jun 28, 2018

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