Let f(x) = \( \left\{ \begin{array}{cl} -x + 3 & \text{if } x \le 0, \\ 2x - 5 & \text{if } x > 0. \end{array} \right.\)

How many solutions does the equation f(f(x)) = 4 have?

IMVERYBADATMATHHELP Aug 13, 2023

#1**0 **

Let’s consider the function f(x)=x2−3. If we want to find the solutions to the equation f(f(x))=4, we can substitute the definition of f(x) into the equation to get:

f(f(x))=(x2−3)2−3=4

Expanding the left side, we get:

x4−6x2+6=4

Subtracting 4 from both sides, we have:

x4−6x2+2=0

This is a quadratic equation in x2, so we can solve for x2 using the quadratic formula:

x2=2(1)−(−6)±(−6)2−4(1)(2)=26±32=3±22

Since x2 must be non-negative, we have two possible values for x2: 3+22 and 3−22. Taking the square root of both sides, we get four possible values for x:

x=±3+22 or x=±3−22

However, since 3−22 is negative, the second pair of solutions is not valid. Therefore, the equation f(f(x))=4 has two solutions: x=3+22 and x=−3+22.

SpectraSynth Aug 13, 2023