anonymous
  • anonymous
which points are the best approximations of the relative maximum and minimum of the function f(x)=x^3+3x^2-9x-8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
theser are the answer options HELP!!
amoodarya
  • amoodarya
find f' find roots of f'=0 check them by f' sign table or ,sign of f'' to find out max or min
anonymous
  • anonymous
find f'(x) and then equate to 0 to find the critical points

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More answers

anonymous
  • anonymous
then check f"(x) at each point.
anonymous
  • anonymous
would it be B??
anonymous
  • anonymous
i really need please pls someone
anonymous
  • anonymous
we have not solved yet. you solve t and show ,we will guide you.
anonymous
  • anonymous
i just dont even know where to start..
anonymous
  • anonymous
differentiate with respect to x
anonymous
  • anonymous
\[f \prime \left( x \right)=3x^2+6x+9\]equate it to zero and find values of x
anonymous
  • anonymous
correction it is -9 not +9
anonymous
  • anonymous
so f(x) = 9 when equated to 0??
anonymous
  • anonymous
no \[f \prime \left( x \right)=0~means~3x^2+6x-9=0\] divide by 3
anonymous
  • anonymous
oh so x^2 + 2x - 3 = 0
anonymous
  • anonymous
make factors
anonymous
  • anonymous
x^2+(3-1)x-3=0 x^2+3x-1x-3=0
anonymous
  • anonymous
how do u make factors
anonymous
  • anonymous
3-1=-3 3*-1=-3
anonymous
  • anonymous
correction 3-1=2
anonymous
  • anonymous
coefficient of x^2*constant term is same as the product of two terms (made from coefficient of x)
anonymous
  • anonymous
ok i still dont understand ugh
anonymous
  • anonymous
coefficient of x^2=1 constant term=-3 product=1*-=-3 if the product is negative always subtract something to make the middle term here 2=3-1 product of 3*-1=-3
anonymous
  • anonymous
i am leaving for sometime . i will be back to help you after 5 minutes.
anonymous
  • anonymous
ok
anonymous
  • anonymous
nevermind i figured it out
anonymous
  • anonymous
i am here.
anonymous
  • anonymous
have you found f"(x)
anonymous
  • anonymous
Have you calculated critical points? \[f \prime \prime \left( x \right)=6x+6\]
anonymous
  • anonymous
x(x+3)-1(x+3)=0 (x+3)(x-1)=0 x+3=0,x=-3 or x-1=0,x=1
anonymous
  • anonymous
\[f \prime \prime \left( -3 \right)=?\]

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