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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
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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

 

 

 

cool cool cool

CPhill  Mar 15, 2018

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