If

f(x) = x^2 - 4 if x >= 4

f(x) = x^2 + 4 otherwise,

then for how many values of x is f(f(x)) = 5?

Guest May 3, 2022

#1**0 **

When \(f(x) \geq 4\), \((x^2 - 4 \geq 4 \text{ and }x \geq 4) \text{ or } (x^2 + 4 \geq 4 \text{ and }x < 4)\), which is \(x < 4\text{ or }x \geq 4\), which means f(x) is always greater than or equal to 4.

Then f(f(x)) = (f(x))^2 - 4 by definition of f.

Case 1: x >= 4

\((x^2 - 4)^2 - 4 = 5\\ x^4 - 8x^2 + 16 - 4 - 5 = 0 \\ x^4 - 8x^2 + 7 = 0\\ (x^2 - 7)(x^2 - 1) = 0\\ x = \pm \sqrt 7 \text{ (rej.) or }x = \pm 1 \text{ (rej.)}\)

The roots are rejected since none of them are greater than 4.

Case 2: x < 4

\((x^2 + 4)^2 - 4 = 5\\ x^4 + 8x^2 + 16 - 4 - 5 = 0\\ (x^2 + 1)(x^2 + 7) = 0\\ \text{No real solution}\)

Therefore, there are no value of x that would make f(f(x)) = 5.

MaxWong May 3, 2022