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 $f(x)=2$?

Question

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 $f(x)=2$?

jim_thompson5910 · Answer

set each piece equal to 2 and solve for x

x^2 - 5x - 64 = 2 leads to x = ?? or x = ??

x^2 + 3x - 38 = 2 leads to x = ?? or x = ??

zmudz · Answer

but the answer isn't 4 (I got for the first equation - -6, 11 and for the second equation 5, -8)

freckles · Answer

when you solve f(x)=2 
you have to make sure the x's fit in the inequality thingy 
x^2-5x-64=2 when x<=0 so you your x=-6 will only work for that one because 11 is greater than 0
so one solution so far...
x^2+3x-38=2 when x>0 so your x=5 will only work for that one because -8 is less than 0

so 1+1=2 you have 2 solutions