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

Question

Quizplus · Accepted Answer

Firstly, to find the extreme values, we need to find the critical points of the function. We can do so by taking the derivative and solving for when it equals zero:
$$y' = 3x^2 - 12 = 0 \implies x = \pm 2$$
Now we have the critical values, we can use the Second Derivative Test to determine whether they are local maxima, minima, or neither. 
$$y'' = 6x$$
At $x = -2$, $y'' = -12$ which is less than zero, so this is a local maximum. 
At $x=2$, $y''=12$ which is greater than zero, so this is a local minimum. 
Therefore, the correct choice is A.

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 $y = x ^ { 3 } - 12 x + 2$