SOLUTION: for the curve -36+39x-3x^3, state the nature of the stationery points and location of the points?

Question 1052726: for the curve -36+39x-3x^3, state the nature of the stationery points and location of the points?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
-36+39x-3x^3=f(x)
the derivative is -9x^2+39
Set it equal to 0 and solve
9x^2=39
x^2=39/9
x=+/-sqrt(39)/3
Numerically, this is +/-2.082
Graphically, the y value is -90.12 at x=-2.08 and 18.12 and x=+2.08

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