Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.
A) ; saddle point:
B) ; relative maximum value:
C) ; relative minimum value:

Question

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A) ; saddle point: 
B) ; relative maximum value: 
C) ; relative minimum value:

Quizplus · Accepted Answer

We need to find critical points first. To find these points we need to find the derivative of the given function, and set it equal to zero, then solve for $x$.

$f(x) = x^3 - 3x$
$f'(x) = 3x^2 - 3$

$3x^2 - 3 = 0 \Rightarrow x = \pm 1$

Therefore, there are two critical points: $(-1, 4)$ and $(1,-4)$.

Now we use the second derivative test to classify the nature of each point.

$f''(x) = 6x$, so $f''(-1) = -6$ and $f''(1) = 6$.

At $x=-1$, $f''(-1) < 0$, so it is a saddle point.

At $x=1$, $f''(1) > 0$, so it is a relative minimum value.

Therefore the answer is C, and the explanation shows that the function has only one relative minimum point at $(1,-4)$.

Find the Critical Point(s) of the Function