zmudz
  • zmudz
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 \(f(x)=2\)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
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
  • zmudz
but the answer isn't 4 (I got for the first equation - -6, 11 and for the second equation 5, -8)
freckles
  • freckles
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

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