 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# The Second Derivative Test for Relative Maximum and Minimum - Problem 1

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Share

It is also possible to find relative maxima and minima by using the second derivative of a function. First, calculate the first derivative of the function and find its critical points. Then, calculate the second derivative of the function. Plug in the critical points found into the second derivative. If the result is positive, the graph is concave up at that point, so it is a relative minimum. If the result is negative, the graph is concave down at that point, so it is a relative maximum.

If the second derivative is 0 at a critical point, then the second derivative test has failed, and you must use the first derivative test to determine if that point is a maximum or minimum.

We're using the second derivative test, to find relative max and min. Let's take a look at a problem. Find the relative maxima, and minima of the function of g(x) equals 3x to the fourth minus 20x³ plus 17.

So the first thing we need to do, is find both the first and the second derivative. So g', the first derivative, is going to be 4 times 3, 12x³, minus the derivative of 20x³ is 60x². I'll leave it like that for a moment.

The second derivative is going to be 36x² minus 120x. So we have the first, and the second derivatives. Generally, I like to have the first, and second derivative that's factored. It makes it much easier to find critical points, and inflection points.

So for the second derivative, I'll factor that right away. I can pull out a 123x for both of these terms leaving a 3x from this term, and a -10 from the second term. So that's my factored second derivative.

The first derivative factors pretty nicely too. You can pull a 12x² out of this, and you're left with x, and -5. We're going to need that later. Actually we need it right away, because the next step is to find critical points, so I need the factored form of this.

Now critical points come from the 0's of your derivative. 12x² (x minus 5). Well, we have the first derivative is 0 or undefined. Now this derivative is not going to be undefined anywhere, but it will equal 0 if x equals 0, or x equals 5. These are our two critical points.

Now the way the second derivative test works is, I test each of these points in the second derivative. So here is my second derivative, I'll copy it down; G'(x) is 12x times 3x minus 10. So I'm going to pop in 0. G''(0) is 12 times 0, immediately I know the result's going to be 0.

Now this is a problem, because the second derivative test doesn't say anything about the second derivative equalling 0. It says if the first derivative is 0, and the second derivative is positive. Or if the first derivative is 0, and the second derivative is negative. But there is no conclusion to the second derivative test if the second derivative is 0, we'll come back to this in a second.

Let's try the other point. G''(5) is 12 times 5, times 3 times 5 minus 10. This is going to be 15 minus 10, or 5 times 60, that's 300. All that matters is that it's positive. So the fact that G''(5) is positive, and G'(5) is 0, indicates that we have, and I always like to draw a little picture here. If G'' is positive, the graph's going to be concave up. G'' is 0, so it's got a horizontal tangent. This indicates a relative minima, so G has a relative minimum at x equals 5.

Now we have to figure out what's happening at x equals 0, because the second derivative test failed. So what do you do when that happens? Well, when the second derivative test fails, you've got to use the first derivative test. So go back to the first derivative, which is G''(x) equals 12x² times x minus 5. We still remember our critical points; 0 and 5. Remember that the first derivative test basically says, make a sign chart. Look for changes in the sign of G'.

You know that G' is 0 at these two points. That's how we found them. These are the points that make the derivative 0. So we just test the interval here, here, and here to see whether G' is positive of negative. So let's try like -1. I plug -1 in here, I get 12 times -1², 12 times 1. Then I get -1 minus 5, -6. So a positive times a negative gives me a negative. Then I test a point in here between 0 and 5, Let's try 1. Again 12 times 1 is positive, 1 minus 5, -4, again negative. Positive times negative is negative. Actually we don't need to test the rest. We already know that there is going to be a relative minimum here. We already found that, so we don't need to know what happens afterwards.

This is what we need to see. We see that the first derivative is negative before, and negative after. That means that this function G is decreasing levels off, and then decreases some more. There is neither a relative max, or min at this point. So I'll write that; neither a max nor a min at x equals 0.

So that's what you do when the second derivative test fails to work. You've got to use the first derivative test. Remember the reason we use the second derivative test at all, is that sometimes when it works perfectly, it works much easier than the first derivative test.