Find all critical points of 𝑓(𝑥)=6𝑥^3−18𝑥−32.

Answer 1

Take the derivative, set it equal to zero, and solve. Also, look for any place the derivative does not exist.

Use the power rule to take the derivative. So f '(x) = 3(6)x^(3-1) - 1(18)x^(1-1) - 0 = 18x² - 18

18x² - 18 = 0

18x² = 18

x² = 1

x = 1 and x = -1

There are no values of x where the derivative does not exist so the critical points are x = 1 and x = -1