Find the extreme values of the function and where they occur.
-$$y = x ^ { 3 } - 12 x + 2$$
A) None
B) Local maximum at (2, -14), local minimum at (-2, 18).
C) Local maximum at (-2, 18), local minimum at (2, -14).
D) Local maximum at (0, 0).

Question

Quizplus · Accepted Answer

To find the extreme values, we need to find the critical points of the function. Taking the derivative of the function, we get: 
$$y' = 3x^2 - 12 = 3(x^2-4)$$
Setting this equal to 0, we get critical points at $x=-2$ and $x=2$. To determine whether these are local maxima or minima or neither, we can look at the sign of the second derivative at these points. Taking the derivative of $y'$, we get: 
$$y'' = 6x$$
So $y''<0$ at $x=-2$ and $y''>0$ at $x=2$, indicating that $x=-2$ is a local maximum and $x=2$ is a local minimum. Therefore, the answer is C.

Find the Extreme Values of the Function and Where They y=x3−12x+2y = x ^ { 3 } - 12 x + 2y=x3−12x+2

Find the Extreme Values of the Function and Where They $y = x ^ { 3 } - 12 x + 2$