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?

waffles Mar 15, 2018

#1**+1 **

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

CPhill Mar 15, 2018