f' = 3x^2-9
so, any extrema will be at x = sqrt(3) or -sqrt(3)
f'' = 6x, so it is not zero there.
So, both values are critical values.
Find the relative extrema, if any, of the function. Use the Second Derivative Test if applicable. (If an answer does not exist, enter DNE.)
g(x)=x^(3)-9x
1 answer
1 answer