Find the absolute max and min values of f(x) =6x^3-9x^2-36x+3 on the interval [-2,4]

So far I have taken the derrivative and got 12x^2-18x-36 in which I simplified to 6(2x^2-3x-6) would I then use the quadratic formula?

Question

Find the absolute max and min values of f(x) =6x^3-9x^2-36x+3 on the interval [-2,4]

So far I have taken the derrivative and got 12x^2-18x-36 in which I simplified to 6(2x^2-3x-6) would I then use the quadratic formula?

jim_thompson5910 · Answer

yes your next step is to solve 12x^2-18x-36 = 0 for x to get the critical values

then you do the first derivative test to see which values are local max values (if any exist)

do not forget to check the endpoints as well