Find the extreme values of the function and where they occur.
-$$y = x ^ { 3 } - 3 x ^ { 2 } + 1$$
A) Local minimum at $( 2 , - 3 )$.
B) Local maximum at $( 0,1 )$.
C) None
D) Local maximum at $( 0,1 )$, local minimum at $( 2 , - 3 )$.

Question

Quizplus · Accepted Answer

To find the extreme values of the function, we need to find the critical points. We can find the critical points by taking the derivative of the function and setting it equal to zero:
$$y' = 3x^2 - 6x = 3x(x-2)$$
Setting this equal to zero gives us the critical points at $x=0$ and $x=2$. 
Next, we can use the second derivative test to determine whether these points are local maxima or minima:
$$y'' = 6x - 6$$
At $x=0$, $y''=-6$, so this is a local maximum. At $x=2$, $y''=6$, so this is a local minimum. Therefore, the answer is D, with a local maximum at $(0,1)$ and a local minimum at $(2,-3)$.

Find the Extreme Values of the Function and Where They y=x3−3x2+1y = x ^ { 3 } - 3 x ^ { 2 } + 1y=x3−3x2+1

Find the Extreme Values of the Function and Where They $y = x ^ { 3 } - 3 x ^ { 2 } + 1$