+17 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?
+121 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.
