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zmudz
 one year ago
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 \(f(x)=2\)?
zmudz
 one year ago
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 \(f(x)=2\)?

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1set 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
 one year ago
Best ResponseYou've already chosen the best response.0but the answer isn't 4 (I got for the first equation  6, 11 and for the second equation 5, 8)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1when you solve f(x)=2 you have to make sure the x's fit in the inequality thingy x^25x64=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+3x38=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
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