Find all relative extrema of the function. Use a graphing utility to verify your result. (If an answer does not exist, enter DNE.)

Question

Find all relative extrema of the function. Use a graphing utility to verify your result. (If an answer does not exist, enter DNE.)

y = ln(3 + 9x^2 − x^3)

Answer 1

Finding the critical points requires taking the derivative and finding the zeroes of it.

The chain rule here gives us the derivative of the inside function times the derivative of the outside function with the old argument.

18x-3x² * (1 / [3 + 9x² - x³]) = [18x-3x²] / [3 + 9x² - x³])

Where is this equal to zero? When the numerator is zero. 3x(6-x)=0 at x=0 and x=6

Plugging those values in you get (0, 1.099) and (6, 4.710). There is an endpoint as well, seeing how this is a log function which cannot be less than zero and stay real. 3+9x²-x³=0 at x≈9.03674 so the graph ends there with a vertical asymptote.

Answer 2

y' = (18x - 3x²)/(3 + 9x² - x³); 3 + 9x² - x³ > 0;

18x - 3x² = 0; x = 0 and x = 6 both in domain.

y'' = [(18 - 6x)(3 + 9x² - x³) - (18x - 3x²)²]/(3 + 9x² - x³)² = (54 - 18x - 162x² + 36x³ - 3x⁴)/(3 + 9x² - x³)^2;

^{y''(0) = 54/9 = 6 > 0; at x =0 y has minimum; (0, ln3) local minimum;}

^{y''(6) = - 6/37 < 0; at x = 6 y has local maximum, (6, ln111) local maximum}