Chapter 3 Graphing and - Maximum Minimum Differential calculus provides tests for locating the key features of graphs. Now that we know how to differentiate, we can use this information to assist us in plotting graphs. The signs of the derivative and the second derivative of a function will tell us which way the graph of the function is "leaning" and "bending." Using the derivative to predict the behavior of graphs helps us to find the points where a function takes on its maximum and minimum values. Many interesting word problems requiring the "best" choice of some variable involve searching for such points. In Section 3.1, we study the geometric aspects of continuity. This will provide a useful introduction to graphing. In Section 3.6, we use the ideas of maxima and minima to derive an important theoretical result-the mean value theorem. One consequence of this theorem is a fact which we used in connection with antiderivatives: a function whose derivative is zero must be constant. 3.1 Continuity and the Intermediate Value Theorem If a continuous function on a closed interval has opposite signs at the endpoints, it must be zero at some interior point. In Section 1.2, we defined continuity as follows: "A function f(x) is said to be continuous at x,, if lin,,,o f(x) = f(x,)." A function is said to be continuous on a given interval if it is continuous at every point on that interval. If a function f is continuous on the whole real line, we just say that "f is continuous." An imprecise but useful guide is that a function is continuous when its graph can be drawn "without removing pencil from paper." In Figure 3.1 .l, the curve on the left is continuous at x,, while that on the right is not. Copyright 1985 Springer-Verlag. All rights reserved. "140 Chapter 3 Graphing and Maximum-Minimum Problems a discontinuous curve I 0 I + (right). ," 0 Y X0 I Example 1 Decide where each of the functions whose graphs appear in Fig. 3.1.2 is continuous. Explain your answers. Ftwe 3.1.2. Where are these functions continuous? Solution (a) This function jumps in value at each of the points x, = 0, x, = 2 1, x0 = f2, . , so limx,xo f(x) does not exist at these points and thus f is not continuous there; however, f is continuous on each of the intervals between the jump points. (b) This function jumps in value at x, = - 1 and x, = + 1, and so lim,,,, f(x) and limx,-, f(x) do not exist. Thus f is not continuous at x, = +- 1; it is continuous on each of the intervals (- m, - l), (- 1, l), and (1900). (c) Even though this function has sharp corners on its graph, it is continuous; lim,,xo f(x) = f(x,) at each point x,. (d) Here lim,,, f(x) = 1, so the limit exists. However, the limit does not equal f(1) = 2. Thus f is not continuous at x, = 1. It is continuous on the intervals (- m, 1) and (1, m). A In Section 1.2, we used various limit theorems to establish the continuity of functions that are basic to calculus. For example, the rational function rule for limits says that a rational function is continuous at points where its denominator does not vanish. Example 2 Show that the function f(x) = (x - 1)/3x2 is continuous at x, = 4. Solution This is a rational function whose denominator does not vanish at x, = 4, so it is continuous by the rational function rule. A Example 3 Let g(x) be the step function defined by Show that g is not continuous at x, = 0. Sketch. Copyright 1985 Springer-Verlag. All rights reserved. 3.1 Continuity and the Intermediate Value Theorem 141 Figure 3.1.3. This step function is discontinuous at x,, = 0. Solution The graph of g is shown in Fig. 3.1.3. Since g approaches (in fact, equals) 0 as x approaches 0 from the left, but approaches 1 as x approaches 0 from the right, lim,,,g(x) does not exist. Therefore, g is not continuous at x, = 0. A Example 4 Using the laws of limits, show that iff and g are continuous at x,, so is fg. Solution We must show that limXjxo( fg)(x) = ( fg)(x,). By the product rule for limits, limxjxo[f(x)g(x)l = Ilim,,,of(x>l[limx jxog(x)l.= f(xo)g(xo), since f and g are continuous at x,. But f(x,)g(x,) = (fg)(x,), and so lim,,,o< fg>(x) = ( fg)(xo), as required. A In Section 1.3, we proved the following theorem: f is differentiable at x,, then f is continuous at x,. Using our knowledge of differential calculus, we can use this relationship to establish the continuity of additional functions or to confirm the continuity of functions originally determined using the laws of limits. Example 5 (a) Show that f(x) = 3x2/(x3- 2) is continuous at x, = I. Where else is it continuous? (b) Show thaiflx) = is continuous at x = 0. Solution (a) By our rules for differentiation, we see that this function is differentiable at x, = 1; indeed, x3 - 2 does not vanish at x, = 1. Thus f is also continuous at x, = 1. Similarly, f is continuous at each x, such that xi - 2.5: 0, i.e., at each x, # 3$. (b) This function is the composition of the square root function h(u) = & and the function g(x) = x2+ 2x + 1; f(x) = h(g(x)). Note that g(0) = 1 > 0. Since g is differentiable at any x (being a polynomial), and h is differentia- ble at u = l, f is differentiable at x = 0 by the chain rule. Thus f is continuous at x = 0. A According to our previous discussion, a continuous function is one whose graph never "jumps." The definition of continuity is local since continuity at each point involves values of the function only near that point. There is a corresponding global statement, called the intermediate value theorem, which involves the behavior of a function over an entire interval [a, b]. Let f be continuous1 on [a, b] and suppose that, for some number c, f (a) < c < f (b) or f(a) > c > f (b). Then there is some point x, in (a, b) such that f (x,) = c. ' Our definition of continuity on [a,b] assumes that f is defined near each point 2 of [a,b], including the endpoints, and that lim,, f(x) = f(SZ). Actually, at the endpoints, it is enough to assume that the one-sided limits (from inside the interval) exist, rather than the two-sided ones. Copyright 1985 Springer-Verlag. All rights reserved. 142 Chapter 3 Graphing and Maximum-Minimum Problems In geometric terms, this theorem says that for the graph of a continuous function to pass from one side of a horizontal line to the other, the graph must meet the line somewhere (see Fig. 3.1.4). The proof of the theorem depends on a careful study of properties of the real numbers and will be omitted. (See the references listed in the Preface.) However, by drawing additional graphs like those in Fig. 3.1.4, you should convince yourself that the theorem is reason- able. Figure 3.1.4. The graph off must pierce the horizontal liney = c if it is to get across. Example 6 Show that there is a number x, such that xi - x, = 3. Solution Let f(x) = x5 - x. Then f(0) = 0 and f(2) = 30. Since 0 < 3 < 30, the interme- diate value theorem guarantees that there is a number x, in (0,2) such that f(x,) = 3. (The function f is continuous on [O, 21 because it is a polynomial.) A Notice that the intermediate value theorem does not tell us how to find the number x, but merely that it exists. (A look at Fig. 3.1.4 should convince you that there may be more than one possible choice for x,.) Nevertheless, by repeatedly dividing an interval into two or more parts and evaluating f(x) at the dividing points, we can solve the equation f(xo) = c as accurately as we wish. This method of bisection is illustrated in the next example. @ Example 7 (The method of bisection) Find a solution of the equation x5 - x = 3 in (0,2) to within an accuracy of 0.1 by repeatedly dividing intervals in half and testing each half for a root. Solution In Example 6 we saw that the equation has a solution in the interval (0,2). To locate the solution more precisely, we evaluate f(l) = l5 - 1 = 0. Thus f(1) < 3 < f(2), so there is a root in (1,2). Now we bisect [1,2] into [l, 1.51 and [1.5,2] and repeat: f(1.5) x 6.09 > 3. so there is a root in (1, 1.5); f(1.25) = 1.80 < 3, so there is a root in (1.25, 1.5); f(1.375)% 3.54 > 3, so there is a root in (1.25, 1.375); thus x, = 1.3 is within 0.1 of a root. Further accuracy can be obtained by means of further bisections. (Related techniques for root finding are suggested in the exercises for this section. Other methods are presented in Section 1 1.4.) A There is another useful way of stating the intermediate value theorem (the contrapositive statement).