# Find the relative extrema, if any, of the function. Use the second derivative test, if...

## Question:

Find the relative extrema, if any, of the function. Use the second derivative test, if applicable.

{eq}f(x) = 3x^3 + 1 {/eq}

## Second Derivative Test

If a function has a local maximum or minimum, it will occur when the first derivative is zero. Thus, we can solve for when the first derivative is zero to find the critical points. We then need to use the second derivative test to determine what they correspond to. A negative second derivative at a critical point corresponds to a local maximum, and a positive second derivative at a critical point corresponds to a local minimum.

To find the relative extrema, we first need to find the critical points. We can begin this procedure by finding the first derivative of this function.

{eq}f'(x) = 9x^2 {/eq}

Next, we can set this equal to zero to find these critical points.

{eq}9x^2 = 0\\ x = 0 {/eq}

Since we only have one critical point, we only have the possibility of one extreme value. Let's use the second derivative to classify this point.

{eq}f''(x) = 18x\\ f''(0) = 18(0) =0 {/eq}

When the second derivative test results in the second derivative being zero at a critical point, that means the critical point we found is not an extreme value. Thus, this function does not have any extrema.