Section 5: Applications of Derivatives

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If you have read The Little Prince by Antoine de Saint-Exupéry, then you remember when the Little Prince first appears, he asks the narrator to draw him a sheep. The first sheep the narrator draws is too sickly. The second is, in the Prince's estimation, a ram. The third is too old. Finally, not knowing exactly what the Little Prince wants, the narrator draws a box and tells him that the sheep is inside. And the sheep inside turns out to be just what the Little Prince had in mind.

In your past math classes you have been asked more often than you'd like to recall to draw a graph of this function or that function. And how often have the graphs you've drawn been too steep or too angular or the wrong shape or had some other fatal defect? And now in introductory calculus you are likely to be asked again to draw the graph of this or that function. If only you could just draw a box and assure your instructor that the graph he wants is inside.

But one thing has changed this time about your being asked to draw graphs, and that is the instructor's agenda. What you have learned about limits and derivatives will, if you apply it correctly, supply you with a box that only the graph your instructor asks for will fit into. And that's the point. By observing the calculus properties of functions, you will be able to determine a lot about a function's graph before you ever plot the first point -- things like where it has asymptotes, where it has maximums and minimums, where it has inflection points -- these form the box. And not many wrong answers will fit into that box.

Graph Problem 1

The best thing to do is to jump right in and try one. Let's do

f(x) =

x² + 1 x² - 4x + 3

The graphing problems that appear on homeworks and exams usually give you hints by asking you to identify where the graph is increasing, where it is decreasing, where the critical points are, whether the critical points are maximums or minimums, and where the horizontal and vertical asymptotes are.

The function we have here falls into the category of being a rational function. That is, it's the ratio of two polynomials. When you get one of these, the first thing to do is to try to factor the polynomials -- especially the denominator. Here, the numerator is x² + 1, which cannot be factored. But the denominator, x² - 4x + 3, can be factored into (x - 1)(x - 3). That means that

f(x) =

x² + 1 (x - 1)(x - 3)

is entirely equivalent to the original function. This gives you your first clues about the behavior of this function. Observe that the denominator has zeros at x = 1 and at x = 3. Test the numerator at those same x's. If the numerator does not have zeros at exactly the same x's, then each unmatched zero in the denominator represents a vertical asymptote on the graph. So at this point you would draw two vertical lines, one at x = 1 and the other at x = 3. Because these vertical lines are asymptotic to the function, the graph of the function will, in the region of each asymptote, grow closer and closer to being parallel to the asymptote, but it will never cross the asymptote.

We can find out much more about the function by looking at its derivative. Since f(x) is a ratio, we have to use the quotient rule. The quotient rule instructs us to take the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared in order to find f'(x).

f'(x) =

(x² - 4x + 3)(2x) - (x² + 1)(2x - 4) (x² - 4x + 3)²

When you multiply out the numerator and gather up like terms, you find that the x³ terms cancel and you are left with

f'(x) =

-4x² + 4x + 4 (x² - 4x + 3)²

Observe that the denominator is squared and therefore it is never negative. So you can determine where the derivative is positive or negative solely by analysing the numerator of f'(x). And knowing where f'(x) is positive and negative will tell you where the function is increasing and decreasing respectively.

The best way to find out where f'(x) is positive and negative is to find where it is zero. Again, where it is zero is almost entirely dependent on the numerator. The denominator cannot cause the function to be zero. So we look for where the numerator is zero. To do this we use the quadratic formula. We find that the numerator has zeros at

x =

1 ± 2

√5 2

In decimals, that's approximately x = -0.61808339 and x = 1.61808339. There is only one more step to finding the zeros of f'(x). That is to plug the numerator's zeros (the x's we just found) into the denominator. Each of the numerator's zeros that is not also a denominator zero is a zero of f'(x). In this case neither of the numerator zeros we found cause the denominator to be zero as well. So at both of these x's we have f'(x) = 0. For this reason both of these x's are critical points.

Now we put the critical x's back into f(x) (you'll want a calculator to do this).

   f(-0.61808339)  =   0.2360679

   f(1.61808339)   =  -4.2360679

These two points ( (-0.61808339,0.2360679) and (1.61808339,-4.2360679)) will be points on the graph, so you might as well plot them. Label them as critical points. And that means that the trace of the graph will be horizontal as it passes through each them.

Two critical points divide the domain (that is the set of all possible x's) into three regions. The first is where x < -0.61808339. The second is -0.61808339 < x < 1.61808339. And the third is x > 1.61808339. We need to test just one x in each region to find out what the sign of f'(x) is in the entire region. So, x = -1 falls into the first region. x = 0 falls into the second region. x = 2 falls into the third region.

   f'(-1)  =  -1/16

   f'(0)   =   4/9

   f'(2)   =  -4/1

This means that in the first region, f(x) is decreasing because f'(x) is negative in that entire region. In the second region, f(x) is increasing because f'(x) is positive in that entire region. In the third region, f(x) is decreasing again because again f'(x) is negative for that entire region.

This also tells us which of the critical points is a minimum and which is a maximum. When x is less than the critical point, -0.61808339, f(x) is decreasing, and when x is just a bit greater than that critical point, f(x) is increasing. Think about it. On both sides of the critical point, f(x) must be more than what it is at the critical point. This clearly indicates that this critical point is a minimum. Likewise, when x is greater than the critical point, 1.61808339, f(x) decreasing. In the region where x is less than that critical point, f(x) is increasing. So on both sides of this second critical point, f(x) must be less than what it is at the critical point. This indicates that the second critical point is a maximum.

Finally look at the original f(x) one more time.

f(x) =

x² + 1 x² - 4x + 3

If you take the limit as x goes to either infinity or to minus infinity, notice that the terms that become dominant are the x² in the numerator and the x² in the denominator. All other terms become insignificant compared to these two. So in the limit (in either direction) you end up with x²/x² = 1. This clues you in that the horizontal asymptote is at y = 1, so draw a horizontal line there. As f(x) goes off to infinity you expect that it will grow closer and closer to being parallel to this line without crossing it. Likewise as x goes off to minus infinity.

Here, at last, is the graph itself. Look at what happens at each of the vertical asymptotes (which are shown in brown). In the region of the first one (where x = 1) we determined that f(x) is increasing. This means that as x approaches the asymptote from the left f(x) must head for positive infinity. On the other side of that asymptote f(x) is still increasing. So its trace must return from minus infinity just to the right of that asymptote. Likewise in the region of the second asymptote (where x = 3) we determined that f(x) was decreasing. That means that as x approaches that asymptote from the left, f(x) heads for minus infinity. And f(x) is still decreasing on the other side of the asymptote, so its trace must return from plus infinity just to the right of that asymptote.

If you make all of the observations we have made here and plotted just the two critical points, you can get the correct shape for the rest of the graph without using your calculuator at all beyond plotting those two points.

On the extreme left the trace of f(x) must be near the horizontal asymptote, so f(x) must be near 1. In addition, since f(x) is decreasing in that region and must meet up with the point, (-0.61808339,0.2360679), without crossing the horizontal assymtote, we know that f(x) is less than 1 and greater than 0.2360679 until it gets to the first critical point.
Going through the first critical point the trace of the graph must be horizontal. The graph reaches a local minimum here.
Following the first critical point, f(x) is increasing, so the trace slopes up.
As it approaches the first vertical asymptote, it heads for plus infinity without ever crossing that asymptote.
To the right of the first vertical asymptote, the trace returns from minus infinity and is still increasing until it gets to the second critical point.
Going through the second critical point the trace of the graph must once again be horizontal. The graph reaches a local maximum here.
Following the second critical point, f(x) is decreasing.
As it approaches the second vertical asymptote, the trace heads for minus infinity.
Following the second vertical asymptote, f(x) is still decreasing. So it must be returning from plus infinity.
As x goes to the extreme right, f(x) continues to decrease, but the trace must approach the horizontal asymptote at y = 1 without crossing it.

If you followed all those instructions, you would not have to know any of the points other than the two we've discussed to get the general picture just about right.

Important Pitfall in the Above Method

To decide where f(x) was increasing and decreasing, we divided its domain according to the zeros of f'(x). In this example we ended up with three such regions. Then we analysed just one point in each region to determine the sign of f'(x) in that region. This will work most of the time but not all the time. It works here because the denominator has only simple zeros. That is, the denominator contained a (x - 1) factor and a (x - 3) factor. So it is said to have simple zeros at x = 1 and at x = 3. But if it had contained, for example, a (x - 1)², then we would say that it has a double zero at x = 1, and a double zero is not a simple zero.

The important thing to remember here is that the sign of f'(x) will not change as a result of x passing through a simple zero of the denominator of f(x) (provided the numerator does not also have a zero at the same x). That is why the method we used worked. The same cannot be said about x passing through a double zero of the denominator.

If you did not know this rule, then the safest thing for you to do would be to divide the domain of f(x) into more regions. Again use each zero of f'(x) as a dividing point between regions. But also use each vertical asymptote as a dividing point as well. In this problem that would give you five regions:

x < -0.61808339
-0.61808339 < x < 1
1 < x < 1.61808339
1.61808339 < x < 3
x > 3

Now you would choose a sample point from each of these regions and determine the sign of f'(x) at each of the sample points. You can be absolutely sure, regardless of double or multiple zeros in the denominator, that the sign of f'(x) will persist throughout each region (this is because f'(x) is continuous within each region, so the intermediate value theorem applies. If f'(x) does not cross zero, it cannot change sign). If you chose x = -1, x = 0, x = 3/2, x = 2, and x = 4, you would have the following table:

   f'(-1)   =  -1/16

   f'(0)    =   4/9

   f'(3/2)  =   16/9

   f'(2)    =  -4/1

   f'(4)    =  -44/9

From this table you can see that f'(x) changes sign only at x = -0.61808339 and at x = 1.61808339. In any problem where the denominator's zeros are not simple, you must use the vertical asymptotes to divide the domain into more regions. In the case where the denominator does have only simple zeros (like this problem), you can still divide up the domain into more regions using the vertical asymptotes if you want to. And you will still get the right answer. It's just a small amount of extra work.

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Move on to 6.1 Be Fruitful and Multiply (exponentials and logs)

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