+647 Let f be the piecewise function such that
\( f(x) = \begin{cases} x^2 - 5x - 64, & x \le 0 \\ x^2 + 3x - 38, & x > 0 \end{cases}\)
At how many points x does f(x) = 2?
+129899 For the first function
x^2 - 5x - 64 =2
x^2 - 5x -66 = 0 factor
(x - 11) ( x + 6) = 0
The only solution for the first restriction is x = -6 ⇒ (-6, 2)
For the second function
x^2 + 3x - 38 = 2
x^2 + 3x - 40 =0\
(x + 8) ( x - 5) = 0
The only solution for the second restriction is x = 5 ⇒ (5 , 2)
So....two points
