The First Derivative Test for Relative Maximum and Minimum - Problem 3
We are using the first derivative test to find the relative maxima and minima of a function. And here’s another example. This is a harder one; h(x) equals 4x minus 6 over x² plus 4.
Remember the first step is always find the derivative. So for this derivative, we’ll need the quotient rule. I’ll call this the low part and this is the high part. It's low d high, x² plus 4 times the derivative of the high part, and it's 4. Minus high d low and that’s the derivative of the denominator, which is going to be x². Over the square of what's below.
So we have x² plus 4². I’m going to have to distribute the 4 and the 2x and simplify the numerator. I get 4x² plus 16 then I get minus 4x times 2x, minus 8x². Then minus minus, plus 6 times 2x, plus 12 x, over x² plus 4 quantity squared. I need to simplify and then factor this as much as I can. It does simplify. There’s a common term here x². So I’m going to have minus 4x² plus 12x plus 16. And those, they all have 4 as a factor.
So minus 4x² plus 12x plus 16. And that will be over x² plus 4 quantity squared. Did I get that right? Just double check ,yes. In this numerator, let’s pull out the -4. That will leave me with x² minus 3x minus 4, that’s easier to factor.
And again we still have the same denominator. Then, this is much easier to factor. You’ll have x plus or minus something x plus or minus something. So you’ll have something like this. And now I need factors of -4. And I’m thinking -4 and +1 will work, because they multiply out to -4. I'll get -4x plus x, that’s -3x, just like I need.
So this is perfect. So this is my derivative and I’m going to transcribe it up here. H'(x) equals -4. We have an x minus 4, x plus 1, all over x² plus 4 quantity squared. So the hard part is over, now we have to find critical points. Remember there are two kinds of critical points. One is where the derivative is undefined, and then 2 is where the derivative equals 0.
The derivative of this rational function is only going to be undefined when the denominator is 0. But that’s never going to happen. X² plus 4 is always going to be positive, never 0.
So we are not going to have any of those kinds of critical points, but we will have critical points when the derivative to be equal to 0, when the numerator is equal to 0. So h' equals 0 when the numerator; -4(x minus 4)(x plus 1) equals 0. And so you can see the critical points are going to be x equals 4 and x equals -1. Those are our critical points.
And now third step make a sign chart for h'. I need to locate these critical points, and let’s remember that h' was 0 at these critical points. And the critical points divide the number line to 3 intervals. I need to test a point in each of these intervals, to see whether the derivative is positive or negative. I don’t need actual values. I just need to know whether it's positive or negative.
So for example -2. I could plug that -2 into this function. Remember the denominator is always going to be positive, so I don’t really need to worry about it. Let’s just look at the numerator.
I got a negative here, -2 minus 4. And another negative. -2 plus 1 a negative. Three negatives, that’s going to give me a negative