Subject: Calculus

Second Derivative Test

In the previous article, we discuss about the first derivative test and there we learned how to find the local extrema or the relative extrema of a function which corresponds to the relative maximum and minimum points of the said function. In this article, let me share with you something about second derivative test, how it relates to the first derivative test and its possible applications in a few Calculus and Mathematics problems.

Relative extrema

We know already how to solve for the relative extrema of functions by using the first deriative test. However, the method used in first derivative test in finding for the relative extrema is quite lengthy and laborious because we still have to check for the values of the function in its left and right part and evaluate if it is positive or negative and then blah, blah, blah...whew! That was really long.
Second derivative test, like first derivative test can also be used to determine the relative extrema of functions. But unlike the first test, 2nd derivative test employs method in solving relative extrema in a more rigorous and short way. Here are the steps on how it is done. Suppose that f=f(x) is well-defined function:

1. Find the critical values of the function f. This can be done by equating the first dervative of f to zero and then solve for x.

2. Obtain the second derivative of f. This can also be done by differentiating f twice.

3. Determine the value or values of the second derivative of f at the critical points or point. This is possible by substituting the critical points obtained in step 1 to the function obtained in step 2.

4. If step three yields a postive result, then it corresponds to the relative minimum value of f. If the result is negative then it corresponds to the relative maximum value.

So that's it. As simple as that. Instead of going through that long way in first derivative test, second derivative test gives you the exact same result in a relatively short and smooth way. Let's redo the practice in the previous article on first derivative but this time, we'll use the second derivative test. We shall find the relative extrema of the function

f(x)=3x^2-\frac{3}{4}x^4

First step: Finding the critical values.

Differentiating once, we get

f'(x)=6x-3x^3

Equating to zero and solve for x, we have the values of our critical points as 0, \sqrt{2} and -\sqrt{2}.

Second step: Obtaining the second derivative.

Differentiating the function above we have

f''(x)=6-9x^2

Third step: Determining values of f''(x) at the critical points.

Substituting the crtitical ponts to the second derivative, we get

@x=0

f''(0)=6-9*0^2=6

@x=\sqrt{2}

f''(\sqrt{2})=6-9*\sqrt{2}^2=6-18=-12

@x=-\sqrt{2}

f''(-\sqrt{2})=6-9*(-\sqrt{2})^2=6-18=-12

Fourth step: Evaluation.

At the critical point x=0, the value of the second derivative is positive, thus, it corresponds to the relative minimum value. On the other hand, at critical points x=\sqrt{2} and x=-\sqrt{2}, the second derivative yields a negative result, thus the function is at its relative maximum value at these points.

When you take a look at our previous article on first derivative test, you can see that we get the exact same result.

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