+0

# Help ASAP!

-1
271
2

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
0

Aug 24, 2020
#2
+32772
+1

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