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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 \(f(f(x)) = 5\)

DanielCai Jun 28, 2018

#1**+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

CPhill Jun 28, 2018