I get the derivative f'(x)= 3x^2+14x-3 now I am stuck. I used the quadratic equation and set answer = 0 but dont get the right answer. Could someone walk me through the rest please?
I get the derivative f'(x)= 3x^2+14x-3 now I am stuck. I used the quadratic equation and set answer = 0 but dont get the right answer. Could someone walk me through the rest please?
by the quadratic formula, to get $\displaystyle f'(x) = 0$ we want $\displaystyle x = \frac {-14 \pm \sqrt{14^2 + 4(3)(3)}}6$
i come up with -4.87 and .21 ? its still wrong, am i doing something else wrong?
those are correct. make sure you are not violating some other condition. how many decimal places should you use? are you required to give the exact form? etc etc etc