第3.7节 导数与曲线绘制

3.7 DERIVATIVES AND CURVE SKETCHING

If we compute n values of f(x),

f(x1), f(x2),……f(xn),

We obtain n points through which the curve y=f(x) passes. The first and second derivatives tell us something about the shape of the curve in the intervals between these points and permit a much more accurate plot of the curve. It is especially helpful to know the signs of the first two derivatives.

When the first derivative is positive the curve is increasing from left to right, and when the first derivative is negative the curve is decreasing from left to right.

When the first derivative is zero the curve is horizontal. These facts can be proved as a theorem if we define exactly what is meant by increasing and decreasing (see Figures 3.7.1 and 3.7.2).

Figure 3.7.1

Figure 3.7.2

DEFINITION

A function f is said to be constant on an interval I if:

f(x1) = f(x2) for all x1, x2 in I.

f is increasing on I if :

f(x1) < f(x2) whenever x1 < x2 in I.

f is decreasing on I if:

f(x1) > f(x2) whenever x1 < x2 in I.

THEOREM 1

Suppose f is continuous on I and has a derivative at every interior point of I.

(i) If f ′(x) =0 for all interior points x of I, then f is constant on I.

(ii) If f ′(x) >0 for all interior points x of I, then f is increasing on I.

(iii) If f ′(x) <0 for all interior points x of I, then f is increasing on I.

A proof will be given in the next section.

EXAMPLE 1 The curve y=x³ + x -1 has derivative dy/dx = 3x² + 1.

The derivative is always positive, so the curve is always increasing (Figure 3.7.3).

Figure 3.7.3

Let us now turn to the second derivative. It is the rate of change of the slope of the curve, so it has something to do with the way in which the curve is changing direction. When the second derivative is positive, the slope is increasing, and we would expect the curve to be concave upward, i.e., shaped like a ___. When the second derivative is negative the slope is decreasing, so the curve should be shaped like _______(see Figure 3.7.4).

A precise definition of concave upward or downward can be given by comparing the curve with the chord (straight line segment ) connecting two points on the curve.

DEFINITION

Let f be defined on I. The curve y=f(x) is concave upward on I if for any two points x1<x2 in I and any value of x between x1 and x2, the curve at x is below the chord which meets the curve at x1 and x2.

The curve y=f(x) is concave downward on I if for any two points x1< x2 in I and any value of x between x1 and x2, the curve at x is above the chord which meets the curve at x1 and x2 (see Figure 3.7.5).

Figure 3.7.4

Figure 3.7.5

The next theorem gives the geometric meaning of the sign of the second derivative.

THEOREM 2

Suppose f is continuous on I and f has a second derivative at every interior point of I.

(i) If f ′′(x) > 0 for all interior points x of I, then f is concave upward on I.

(ii) If f ′′(x) < 0 for all interior points x of I, then f is concave downward on I.

We have already explained the intuitive reason for Theorem 2. The proof is omitted. Theorem 1 tells what happens when f ′always has the same sign on an open interval I, while Theorem 2 does the same thing for f ′′. To use these results we need another theorem that tells us that certain functions always have the same sign on I.

THEOREM 3

Suppose is continuous on I, and g(x) ≠ 0 for all x in I.

(i) If g(c) > 0 for at least one c in I, then g(x) > 0 for all x in I.

(ii) If g(c) < 0 for at least one c in I, then g(x) < 0 for all x in I.

The two cases are shown in Figure 3.7.6. We give the proof in the next section.

Figure 3.7.6

Let us show with some simple examples how we can use the first and second derivatives in sketching

curves. The three theorems above and the tests for minima and maxima are all helpful.

EXAMPLE 1 (Continued) y=x³ + x -1. We have

dy/dx is always positive, while d²y/dx² = 0 at x = 0. We make a table of values for y and its first two derivatives at x=0 and at a point to the right and left side of 0.

With the aid of Theorems 1-3, we can draw the following conclusions:

(a) dy/dx >0 and the curve is increasing for all x.

(b) d²y/dx²<0 for x<0; concave downward.

(c) d²y/dx² >0 for x >0; concave upward.

At the point x=0, the curve changes from concave downward to concave upward. This is called a point of inflection.

To sketch the curve we first plot the three values of y shown in the table, then sketch the slope at these points as shown in Figure 3.7.7, then fill in a smooth curve, which is concave downward or upward as required.

Figure 3.7.7

EXAMPLE 2 Sketch the curve y=2x - x².

We see that dy/dx = 0 when x = 1, a critical point. d²y/dx²<0 is never zero because it is constant.

We make a table of values including the critical point x= 1 and points to the right and left of it.

CONCLUSIONS

(a) dy/dx > 0 for x < 1; increasing.

(b) dy/dx < 0 for x >1; decreasing.

(c) d²y/dx²<0 <0 for all x; concave downward.

(d) dy/ dx = 0, d²y/dx² < 0 at x=1; maximum.

The curve is shown in Figure 3.7.8.

In general a curve y = f(x) may go up and down several times. To sketch it we need to determine the intervals on which it is increasing or decreasing, and concave upward or downward. Here are some things which may happen at the endpoints of these intervals.

Figure 3.7.8

DEFINITION

Let c be an interior point of I.

f has a local maximum at c if (c) ≥ f(x) for all x in some open interval (a0, b0) containing c.

f has a local minimum at c if (c) ≤ f(c) for all x in some open interval (a0,b0) containing c.

(The interval (a0, b0) may be only a small subinterval of I.)

f has a point of inflection at c if f changes from one direction of concavity to the other at c.

The definitions are illustrated in Figure 3.7.9.

Figure 3.7.9

We may now describe the steps in sketching a curve. We shall stick to the simple case where f and its first two derivatives are continuous on a closed interval [a,b], and either are never zero or are zero only finitely many times. (Curve plotting in a more general situation is discussed in Chapter 5 on limits.)

Step 1 Compute dy/dx and d²y/dx².

Step 2 Find all points where dy/dx = 0 and all points where d²y/dx² = 0.

Step 3 Pick a few points

a = x0, x1,x2,……xn = b

In the interval [a, b]. They should include both endpoints, all points where the first or second derivative is zero, and at least one point between any two consecutive zeros of dy/dx or d² y/dx².

Step 4 At each of the points x0,……xn, compute the values of y and dy/dx and determine the sign of

d²y/dx². Make a table.

Step 5 From the table draw conclusions about where y is increasing or decreasing, where y has a local

maximum or minimum, where the curve is concave upward or downward, and where it has a point

of inflection. Use theorems 1-3 of this section and the tests for maxima and minima.

Step 6 Plot the values of y and indicate slopes from the table. Then connect them with a smooth curve

which agrees with the conclusions of Step 5.

EXAMPLE 3 y = x4/2 - x2, -2 ≤ x ≤ 2.

Step 1 dy /dx = 2x3 - 2x. d2y/dx2 = 6x2 - 2.

Step 2 dy/dx = 0 at x = -1,0,1.

Step 3 d2y/dx2 = 0 at

Step 4

Step 5 We indicate the conclusions schematically in Figure 3.7.10.

Figure 3.7.10

Step 6 The curve is W- shaped, as shown in Figure 3.7.11.

Figure 3.7.11

PROBLEMS FOR SECTION 3.7

Sketch each of the curves given below by the six-step process explained in the text. For each curve, give a

table showing all the critical points, local maxima and minima, intervals on which the curve is increasing or

decreasing, points of inflection, and intervals on which the curve is concave upward or downward.

1 y=x²+2, -2 ≤ x ≤ 2 2 y=1 - x², -2 ≤ x ≤ 2

3 y=x² - 2x, -2 ≤ x ≤ 2 4 _______________

5 y=2x² - 4x + 3, 0≤ x ≤ 2 6 y= -x² - 2x + 6, -4 ≤ x ≤ 0

7 y=x4, -2 ≤ x ≤ 2 8 y=x5, -2 ≤ x ≤ 2

9 y=x³+x²+x, -2 ≤ x ≤ 2

10 y=x³+x² - x, -2 ≤ x ≤ 2

11 __________________

12 y= -x³+12x -12, -3 ≤ x ≤ 3

13 y=x4+4x³+2, -4 ≤ x ≤ 2

14 _____________

15 ____________

16 y=x2(x-2)², -1 ≤ x ≤ 3

17 __________

18 __________

19 __________

20__________

33 y=sin x cos x, 0__ x___ 2__ 34 y=sin x + cos x, 0__ x___ 2__

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37 y= tan x, -π_ /3 x /3 38 y=1/cos x, -_ /3 x _/3

39 y= e-x, -2 x 2 40 y= e(1/2)x, -2 x 2

41 y = 1n x, 1/e x e 42 y = 1n x, 1/e x e

43 y = x e -x, -1 x 3 44 y = x -ex, -2 x 2

45 y = x1n x, e-2 x e 46 y = x - 1n x, e-2 x e

47 y = x e, -3 x 1 48 y=e-x2, -2 x 2

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