We can use calculus to analyze the graphs of differentiable functions of many kinds, not only polynomial functions. Let’s look at a continuous, differentiable function that lends itself nicely to analysis by differentiation: the sine function from trigonometry.
Analyzing the curve
Figure 9-6 is a graph of the sine function over the limited domain where x is restricted to the open interval between, but not including, 0 and 2p. Remember that the x-values are in radi-ans, not in degrees!
We’re familiar enough with trigonometry to know that this curve, with the domain restricted as shown in Fig. 9-6, has an absolute maximum at the point (x,y ) = (p /2,1). It has an absolute minimum at the point (x,y ) = (3p /2,−1). Let’s verify these two facts mathemati-cally. The first derivative is
dy /dx (sin x)= cos x
Graph of the Sine Function 147
0 1
–1
x y
0 p/2 p 3p/2 2p
dy/dx= 0
Absolute minimum dy/dx= 0
Absolute maximum
Inflection point d2y/dx2= 0
Figure 9-6 Graph of y= sin x for 0 < x < 2p. On the x axis, each division represents p/6 unit. On they axis, each division represents 1/5 unit.
and the second derivative is
d 2y /dx 2 (sin x ) = d /dx (cos x ) = − sinx
At the point where x= p /2, the first derivative is the cosine of p/2. That’s 0. The second derivative is the negative of the sine of p /2. That’s −1. The slope is 0 and the curve is concave downward, meaning that we have a maximum. At the point where x= 3p /2, the first deriva-tive is the cosine of 3p /2. That’s 0. The second derivaderiva-tive is the negaderiva-tive of the sine of 3p /2.
That’s 1. The slope is 0 and the curve is concave upward, telling us that we have a minimum.
We can look at the graph and immediately see that these extrema are absolute. If we were to include multiple cycles of the wave (or even infinitely many, by letting the domain extend over the entire set of real numbers), then these extrema would be local.
It looks like the curve in Fig. 9-6 has an inflection point at (x,y ) = (p,0). If we want to be certain, we must look at the second derivative, which is the negative of the sine of p. That’s 0, so we know that (x,y ) = (p,0) is indeed a point of inflection. The curve is concave downward to the left of the point, and concave upward to the right of it.
Are you confused?
All of the curves we’ve seen in this chapter have represented differentiable functions. “But,” you ask,
“what happens if a function has points or intervals where it’s nondifferentiable?” In cases of that sort, we can usually graph the curve, but we can’t use differentiation to evaluate it at the nondifferentiable points or within the nondifferentiable intervals. Even if a function clearly has an extremum at a certain point, we can’t verify this fact with differentiation if the function is nondifferentiable at that point. A good example is the absolute-value function y= |x|. This has an absolute minimum at the point (x,y ) = (0,0), but because the function isn’t differentiable at that point, we can’t use derivatives to verify that fact.
Here’s a challenge!
Sketch a graph of the square of the sine function for 0 < x < 2p. (It’s all right to use a calculator, plot numerous points, and then connect them all by curve fitting to obtain this sketch.) Determine the x-values of the extrema and the inflection points. Verify these results using the calculus techniques we’ve learned.
Solution
Figure 9-7 is a sketch of this function. Symbolically, we can denote the function as either y = (sin x )2
or
y= sin2x In Chap. 7, we found that
d /dx sin2x= 2 sin x cos x
We can differentiate again, obtaining the second derivative, using the multiplication-by-constant rule and the product rule:
d 2/dx 2 (sin2x ) = d /dx (2 sin x cos x ) = 2 [d /dx (sin x cos x )]
= 2 [(d /dx sin x ) cos x+ (d /dx cos x ) sin x] = 2 [cos x cos x+ (− sinx sin x )]
= 2 (cos2x− sin2x ) = 2 cos2x− 2 sin2x
To find the extrema, we must figure out all the x-values where the first derivative is 0. That means we must solve the equation
2 sin x cos x= 0
This is satisfied whenever sin x= 0 or cos x= 0. We’ve restricted x to positive values less than 2p, so we have
sinx= 0 when x=p and
cosx= 0 when x=p /2 or x= 3p /2
Graph of the Sine Function 149
1
–1 y
0 p/2 p 3p/2 2p
0 x
dy/dx= 0
Absolute minimum dy/dx= 0
Local maxima
Inflection points d2y/dx2= 0
Figure 9-7 Graph of y = sin2x for 0 < x < 2p. On the x axis, each division represents p/6 unit. On the y axis, each division represents 1/5 unit.
When we look at Fig. 9-7, it appears that the extrema for x=p /2 and x= 3 p /2 are maxima. We can ver-ify that these points are maxima by checking the values of the second derivatives. For x=p /2, we have
d 2/dx 2 (sin2x ) = 2 cos2x− 2 sin2x = 2 (cos p /2)2− 2 (sin p /2)2
= 2 · 02− 2 · 12 = −2
The fact that the second derivative is negative indicates that the curve is concave downward, so this extre-mum is a maxiextre-mum. In the case of x= 3p/2, we have
d 2/dx 2 (sin2x ) = 2 cos2x− 2 sin2x = 2 (cos 3p /2)2− 2 (sin 3p /2)2 = 2 · 02− 2 · (−1)2 = −2
Again, the curve is concave downward, so we know that the function attains another maximum at this point.
Are you astute?
“Wait,” you say. “We’ve found two maxima, all right. But are they local, or is one of them absolute?” If the y-values are the same for both points, then they’re both local maxima. But if the y-values are different, then the point with the larger y-value is the absolute maximum. We’d better check this out! When we plug in x=p /2 to the original function, we get
y = sin2x = (sinp /2)2 = 12 = 1
When we plug in x = 3p /2 to the original function, we get
y = sin2x = (sin 3p /2)2 = (−1)2 = 1 Now we know that these two maxima are the same, so they’re both local.
Back to the challenge!
Let’s return to the “challenge” we’ve been dealing with. There appears to be an extremum at the point where x=p. Figure 9-7 suggests that it’s a minimum. To verify, let’s plug in p for x and see what we get for the second derivative:
d 2/dx 2 (sin2x ) = 2 cos2x− 2 sin2x = 2 (cos p)2− 2 (sin p)2 = 2 · (−1)2− 2 · 02= 2
It’s positive, all right! That means the curve is concave upward at this point, so it must represent a minimum.
Within the span of x-values we’ve allowed here, it’s the absolute minimum, because we’ve left x= 0 and x= 2p out of the domain. (If we had left those points in the domain, they’d represent local minima, as would the point where x=p. You can verify, if you like, that all three of these minima would be the same.)
The inflection points occur where the second derivative is equal to 0. It appears that there are four such points in the curve of Fig. 9-7, but it’s difficult to tell exactly where they are. We must solve the following equation to find the x-values:
2 cos2x− 2 sin2x= 0
Dividing through by 2, we get
cos2x− sin2x= 0 Adding sin2x to each side gives us
cos2x= sin2x
We can take the square root of both sides here, keeping in mind that we have to account for the positive and negative values. When we do that, we get
± cosx= ± sinx
We know that the inflection points don’t occur where either the sine or the cosine are equal to 0. Those situations represent the extrema. We’ve already determined that the second derivatives are nonzero at those points. Knowing that the cosine isn’t 0 at any of the inflection points, we can divide each side of the above equation by ±cosx to get
±1= (± sinx ) / (± cosx )
From trigonometry, we remember that the sine divided by the cosine is equal to the tangent. This means we can simplify the above equation to
± tan x= ±1 which tells us that
x= Arctan (±1)
This gives us four values within the domain we’ve allowed, which is 0 <x< 2p. Those values are x=p /4
x= 3p /4 x= 5p /4 x= 7p /4
Here’s an extra-credit challenge!
Determine the y-values of each of the inflection points on the curve shown in Fig. 9-7. Then write down the complete (x,y ) coordinates of all four points.
Solution
You’re on your own. That’s why you get extra credit!
Graph of the Sine Function 151
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems. Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. Determine the coordinates of the inflection point in the graph of the function y= 3x 3+ 3x 2−x− 7
2. What is the slope of the curve representing the function stated in Prob. 1 at the inflection point?
3. In what range of x-values is the graph of the function stated in Prob. 1 concave upward?
In what range of x-values is it concave downward?
4. Determine the extremum in the graph of the function y= 4x 2− 7 Is this an absolute maximum or an absolute minimum?
5. Look again at Fig. 9-4. What is the equation of a line tangent to the curve, and passing through the point (2,0)?
6. Consider the cubic function
y=x 3+ 3x 2− 3x+ 4
Without drawing the graph, determine the coordinates of the inflection point. Then calculate the slope of the curve at the inflection point.
7. Evaluate the function stated in Prob. 6 for x= 100. We choose x= 100 because it’s much greater than the absolute values of any of the coefficients. If we get a large
positive value for y, we’ll know that the curve trends upward overall, like
Fig. 9-2A, C, or E. If we get a large negative value for y, we’ll know that the curve trends downward overall, like Fig. 9-2B, D, or F. Note again the slope of the function at the inflection point. We found that slope in the solution to Prob. 6.
From all this information, identify which of the six general contours from Fig. 9-2 applies to this curve.
8. Consider again the function we worked with in the solutions to Probs. 6 and 7. Find they-intercept of the curve. Are there any points where the slope is 0? If so, determine the coordinates of those points. Finally, on the basis of all the information we’ve gathered about this function, sketch its graph.
9. Determine the slope, at the inflection point, of the curve shown in Fig. 9-6 for the function
y = sin x
10. Determine the slope, at each of the four inflection points, of the curve shown in Fig. 9-7 for the function
y = sin2x
Practice Exercises 153
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