Rolle s Theorem. THEOREM 3 Rolle s Theorem. x x. then there is at least one number c in (a, b) at which ƒ scd = 0.

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1 4.2 The Mean Value Theorem The Mean Value Theorem f '(c) 0 f() We know that constant functions have zero derivatives, ut could there e a complicated function, with man terms, the derivatives of which all cancel to give zero? What is the relationship etween two functions that have identical derivatives over an interval? What we are reall asking here is what functions can have a particular kind of derivative. These and man other questions we stud in this chapter are answered appling the Mean Value Theorem. To arrive at this theorem we first need Rolle s Theorem. 0 a c f '(c ) 0 (a) f '(c 2 ) 0 f '(c 3 ) 0 0 a c c 2 c 3 () f() FIGURE 4.0 Rolle s Theorem sas that a differentiale curve has at least one horizontal tangent etween an two points where it crosses a horizontal line. It ma have just one (a), or it ma have more (). Rolle s Theorem Drawing the graph of a function gives strong geometric evidence that etween an two points where a differentiale function crosses a horizontal line there is at least one point on the curve where the tangent is horizontal (Figure 4.0). More precisel, we have the following theorem. THEOREM 3 Rolle s Theorem Suppose that = ƒsd is continuous at ever point of the closed interval [a, ] and differentiale at ever point of its interior (a, ). If ƒsad = ƒsd, then there is at least one numer c in (a, ) at which ƒ scd = 0. Proof Being continuous, ƒ assumes asolute maimum and minimum values on [a, ]. These can occur onl

2 256 Chapter 4: Applications of Derivatives HISTORICAL BIOGRAPHY Michel Rolle (652 79). at interior points where ƒ is zero, 2. at interior points where ƒ does not eist, 3. at the endpoints of the function s domain, in this case a and. B hpothesis, ƒ has a derivative at ever interior point. That rules out possiilit (2), leaving us with interior points where ƒ =0 and with the two endpoints a and. If either the maimum or the minimum occurs at a point c etween a and, then ƒ scd = 0 Theorem 2 in Section 4., and we have found a point for Rolle s theorem. If oth the asolute maimum and the asolute minimum occur at the endpoints, then ecause ƒsad = ƒsd it must e the case that ƒ is a constant function with ƒsd = ƒsad = ƒsd for ever H [a, ]. Therefore ƒ sd = 0 and the point c can e taken anwhere in the interior (a, ). The hpotheses of Theorem 3 are essential. If the fail at even one point, the graph ma not have a horizontal tangent (Figure 4.). f() f() f() a (a) Discontinuous at an endpoint of [a, ] a 0 a 0 () Discontinuous at an interior point of [a, ] (c) Continuous on [a, ] ut not differentiale at an interior point FIGURE 4. There ma e no horizontal tangent if the hpotheses of Rolle s Theorem do not hold. ( 3, 2 3) EXAMPLE The polnomial function Horizontal Tangents of a Cuic Polnomial ( 3, 2 3) FIGURE 4.2 As predicted Rolle s Theorem, this curve has horizontal tangents etween the points where it crosses the -ais (Eample ). graphed in Figure 4.2 is continuous at ever point of [-3, 3] and is differentiale at ever point of s -3, 3d. Since ƒs -3d = ƒs3d = 0, Rolle s Theorem sas that ƒ must e zero at least once in the open interval etween a = -3 and = 3. In fact, ƒ sd = 2-3 is zero twice in this interval, once at = -23 and again at = 23. EXAMPLE 2 Show that the equation has eactl one real solution. ƒsd = Solution of an Equation ƒsd = = 0 Solution Let Then the derivative = ƒsd = ƒ sd =

3 4.2 The Mean Value Theorem 257 (, 3) (, 5) FIGURE 4.3 The onl real zero of the polnomial = is the one shown here where the curve crosses the -ais etween - and 0 (Eample 2). is never zero (ecause it is alwas positive). Now, if there were even two points = a and = where ƒ() was zero, Rolle s Theorem would guarantee the eistence of a point = c in etween them where ƒ was zero. Therefore, ƒ has no more than one zero. It does in fact have one zero, ecause the Intermediate Value Theorem tells us that the graph of = ƒsd crosses the -ais somewhere etween = - (where = -3) and = 0 (where = ). (See Figure 4.3.) Our main use of Rolle s Theorem is in proving the Mean Value Theorem. The Mean Value Theorem The Mean Value Theorem, which was first stated Joseph-Louis Lagrange, is a slanted version of Rolle s Theorem (Figure 4.4). There is a point where the tangent is parallel to chord AB. THEOREM 4 The Mean Value Theorem Suppose = ƒsd is continuous on a closed interval [a, ] and differentiale on the interval s interior (a, ). Then there is at least one point c in (a, ) at which = ƒ scd. () 0 a f() Slope f'(c) A c Tangent parallel to chord Slope B f() f(a) a FIGURE 4.4 Geometricall, the Mean Value Theorem sas that somewhere etween A and B the curve has at least one tangent parallel to chord AB. A(a, f(a)) a f() B(, f()) FIGURE 4.5 The graph of ƒ and the chord AB over the interval [a, ]. Proof We picture the graph of ƒ as a curve in the plane and draw a line through the points A(a, ƒ(a)) and B(, ƒ()) (see Figure 4.5). The line is the graph of the function gsd = ƒsad + (point-slope equation). The vertical difference etween the graphs of ƒ and g at is hsd = ƒsd - gsd Figure 4.6 shows the graphs of ƒ, g, and h together. The function h satisfies the hpotheses of Rolle s Theorem on [a, ]. It is continuous on [a, ] and differentiale on (a, ) ecause oth ƒ and g are. Also, hsad = hsd = 0 ecause the graphs of ƒ and g oth pass through A and B. Therefore h scd = 0 at some point c H sa, d. This is the point we want for Equation (). To verif Equation (), we differentiate oth sides of Equation (3) with respect to and then set = c: ƒ scd = which is what we set out to prove. = ƒsd - ƒsad - h sd = ƒ sd - h scd = ƒ scd - 0 = ƒ scd -, s - ad s - ad. Derivative of Eq. (3) p p with = c h scd = 0 Rearranged (2) (3)

4 258 Chapter 4: Applications of Derivatives HISTORICAL BIOGRAPHY f() B Joseph-Louis Lagrange (736 83) A g() h() 2, a h() f() g() FIGURE 4.6 The chord AB is the graph of the function g(). The function hsd = ƒsd - gsd gives the vertical distance etween the graphs of ƒ and g at. 0 FIGURE 4.7 The function ƒsd = 2-2 satisfies the hpotheses (and conclusion) of the Mean Value Theorem on [-, ] even though ƒ is not differentiale at - and B(2, 4) 2 The hpotheses of the Mean Value Theorem do not require ƒ to e differentiale at either a or. Continuit at a and is enough (Figure 4.7). EXAMPLE 3 The function ƒsd = 2 (Figure 4.8) is continuous for 0 2 and differentiale for Since ƒs0d = 0 and ƒs2d = 4, the Mean Value Theorem sas that at some point c in the interval, the derivative ƒ sd = 2 must have the value s4-0d>s2-0d = 2. In this (eceptional) case we can identif c solving the equation 2c = 2 to get c =. A(0, 0) (, ) FIGURE 4.8 As we find in Eample 3, c = is where the tangent is parallel to the chord. Distance (ft) s 2 s f(t) 5 Time (sec) (8, 352) At this point, the car s speed was 30 mph. FIGURE 4.9 Distance versus elapsed time for the car in Eample 4. t A Phsical Interpretation If we think of the numer sd>sd as the average change in ƒ over [a, ] and ƒ scd as an instantaneous change, then the Mean Value Theorem sas that at some interior point the instantaneous change must equal the average change over the entire interval. EXAMPLE 4 If a car accelerating from zero takes 8 sec to go 352 ft, its average velocit for the 8-sec interval is 352>8 = 44 ft>sec. At some point during the acceleration, the Mean Value Theorem sas, the speedometer must read eactl 30 mph (44 ft>sec) (Figure 4.9). Mathematical Consequences At the eginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollar of the Mean Value Theorem provides the answer. COROLLARY Functions with Zero Derivatives Are Constant If ƒ sd = 0 at each point of an open interval (a, ), then ƒsd = C for all H sa, d, where C is a constant.

5 4.2 The Mean Value Theorem C C 2 C C 0 C 2 C 2 0 Proof We want to show that ƒ has a constant value on the interval (a, ). We do so showing that if and 2 are an two points in (a, ), then ƒs d = ƒs 2 d. Numering and 2 from left to right, we have 6 2. Then ƒ satisfies the hpotheses of the Mean Value Theorem on [, 2 ]: It is differentiale at ever point of [, 2 ] and hence continuous at ever point as well. Therefore, ƒs 2 d - ƒs d 2 - = ƒ scd at some point c etween and 2. Since ƒ =0 throughout (a, ), this equation translates successivel into ƒs 2 d - ƒs d 2 - = 0, ƒs 2 d - ƒs d = 0, and ƒs d = ƒs 2 d. At the eginning of this section, we also asked aout the relationship etween two functions that have identical derivatives over an interval. The net corollar tells us that their values on the interval have a constant difference. COROLLARY 2 Functions with the Same Derivative Differ a Constant If ƒ sd = g sd at each point in an open interval (a, ), then there eists a constant C such that ƒsd = gsd + C for all H sa, d. That is, ƒ - g is a constant on (a, ). 2 FIGURE 4.20 From a geometric point of view, Corollar 2 of the Mean Value Theorem sas that the graphs of functions with identical derivatives on an interval can differ onl a vertical shift there. The graphs of the functions with derivative 2 are the paraolas = 2 + C, shown here for selected values of C. Proof At each point H sa, d the derivative of the difference function h = ƒ - g is h sd = ƒ sd - g sd = 0. Thus, hsd = C on (a, ) Corollar. That is, ƒsd - gsd = C on (a, ), so ƒsd = gsd + C. Corollaries and 2 are also true if the open interval (a, ) fails to e finite. That is, the remain true if the interval is sa, q d, s - q, d, or s - q, q d. Corollar 2 plas an important role when we discuss antiderivatives in Section 4.8. It tells us, for instance, that since the derivative of ƒsd = 2 on s - q, q d is 2, an other function with derivative 2 on s - q, q d must have the formula 2 + C for some value of C (Figure 4.20). EXAMPLE 5 Find the function ƒ() whose derivative is sin and whose graph passes through the point (0, 2). Solution Since ƒ() has the same derivative as gsd = -cos, we know that ƒsd = -cos + C for some constant C. The value of C can e determined from the condition that ƒs0d = 2 (the graph of ƒ passes through (0, 2)): The function is ƒsd = -cos + 3. ƒs0d = -cos s0d + C = 2, so C = 3. Finding Velocit and Position from Acceleration Here is how to find the velocit and displacement functions of a od falling freel from rest with acceleration 9.8 m>sec 2.

6 260 Chapter 4: Applications of Derivatives We know that (t) is some function whose derivative is 9.8. We also know that the derivative of gstd = 9.8t is 9.8. B Corollar 2, std = 9.8t + C for some constant C. Since the od falls from rest, s0d = 0. Thus 9.8s0d + C = 0, and C = 0. The velocit function must e std = 9.8t. How aout the position function s(t)? We know that s(t) is some function whose derivative is 9.8t. We also know that the derivative of ƒstd = 4.9t 2 is 9.8t. B Corollar 2, sstd = 4.9t 2 + C for some constant C. If the initial height is ss0d = h, measured positive downward from the rest position, then 4.9s0d 2 + C = h, and C = h. The position function must e sstd = 4.9t 2 + h. The ailit to find functions from their rates of change is one of the ver powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.