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The First Derivative Test for Relative Maximum and Minimum - Problem 2

Teacher/Instructor 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.

The relative maxima and minima are critical points of a function. So, as we did to find critical points, to find the relative maxima and minima, first calculate the derivative of the function. Then, find the points at which the derivative is either 0 or undefined. Then, find where the function is increasing and decreasing. A critical point is a maxima if the function changes from increasing to decreasing at that point, and a minima if the function changes from decreasing to increasing at that point.

We are looking at the relative maximum and minimum of a function. Let’s try another example. We want to find the relative max and min of g of x equals x times the quantity 5minus x to the 2/3. This is a trickier function and we’ll need the product rule to differentiate. And that’s the first thing we do is differentiate a functions. So let’s start with product rule. It's the first times the derivative of the second.

The derivative of the second is going to require the chain rule. The 2/3 will come out in front, 5 minus x. So 1 less than 2/3. 2/3 minus 1 is -1/3. And I have to multiply by the derivative of what’s inside which is -1. Don’t forget that. So use the chain rule here. Then plus the second times the derivative of the first. So 5 minus x to the 2/3 and the derivative of the first. The derivative of x is just 1.

And I have to simplify this a lot for it will be useful for me to find critical points. So first let me observe that I have a fraction here. 5 minus x to the -1/3. That’s the same as 1 over 5 minus x to the positive 1/3, and 3 is also on the denominator. So make a fraction of this.

I have 3 times 5 minus x to the 1/3 and in the numerator I have minus 1 times 2, -2x. So that’s this term written in fraction form. This term, 5 minus x to the 2/3, it can also be thought as a fraction, but it doesn’t have the same denominator. I have to give it the same denominator so I can combine these to a single term.

That’s really important because, we will need to do that to find critical points. The denominator I need is 3 times 5 minus x to the 1/3. To get that I have to multiply the top and bottom, of this term, by 3 times 5 minus x to the 1/3.

Let’s see what that does for us. First of all, we will get the same denominator in both terms. This term is really going to be unchanged. -2x and then here we have that denominator 3, 5x to the 1/3. What do we have on the top? 5 minus x to the 2/3 times 5 minus x to the 1/3 is just 5 minus x. You add the exponents. Same bases, you add the exponents you get 1. So it's 3 times 5 minus x. Let's see if we can combine this in a single step. This is 15 minus 3x minus 2x that’s 15 minus 5x all over 3 times 5 minus x to the 1/3.

That’s our derivative, simplify. This will be really easy to find critical points. Let’s take this up to the next board. G'; 15 minus 5x over 3 times 5 minus x to the 1/3.

Remember the critical points are points with the derivative equals 0 or is undefined. It will have one of these kind of critical point here. The derivative is undefined at x equals 5. So g'(5) is undefined. And that means x equals 5 is a critical point. We also have the derivative equals 0 when x equals 3. Notice the numerator will be 0 when x equals 3. So g'(3) is 0. So that means x equals 3 is another critical point.

Now you’ve identified one of the critical points, then you just need to make sign chart for g'. The sign chart will tell us whether the derivative is positive or negative. We need to know that in order to decide whether these critical points represent relative maxes or mins.

So I’m going to put the critical points 3 and 5 on my sign chart. Let’s remember that g' is 0 at x equals 3. But undefined at x equals 5. And these critical points divide the number line into 3 intervals. I need to test points in each interval to see whether g' is positive or negative. So I will test the point over here, let’s say 0. Plug in 0 into the derivative, I get on top 15, a positive number and here 5 minus 0, another positive number.

I get a positive over a positive, so the derivative is positive. Then I’ll try 4. 4 is going to give me 15 minus 20, a negative number on top. 5 minus 4, positive, negative over positive will be negative. And then finally, to the right of 5, I’ll try 6. 15 minus 5 to 6 minus 30, that’s negative. 5 minus 6 also negative, raised to the 1/3 power still negative.

We have a negative over a negative, positive. So the idea here is we just want to know where g' is positive or negative, because a change from positive to negative, indicates a relative max. Now if you don’t see that right away again, the trick that you can use is to draw a little line. The positive slope means that g is increasing. So draw an increasing line.

And the negative slope means that g is decreasing. And here these slanted lines I’m drawing that represent what g is doing not g'. So g goes from increase to decreasing and that means we have a relative max at x equals 3. So I will write g has a relative max at x equals 3. And this is by the first derivative test.

Then we have that g is increasing here. So it's going from decreasing to increasing. You see that there is a relative mean. Even though the derivative is undefined, there’s a relative mean at x equals 5.

Now remember that this function may be undefined at 5, or the derivative maybe undefined at 5, but the original function, g, was defined at 5. It’s continuous there and so something happens there and it does have a relative mean. So it's important to take in consideration, critical points where the derivative is undefined don’t forget to look for those.

In the end our answer is, there’s a relative max at x equals 3, a relative min at x equals 5.

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