Learning Outcomes
- Explain the meaning of Rolle’s theorem
- Describe the significance of the Mean Value Theorem
- State three important consequences of the Mean Value Theorem
Rolle’s Theorem
Informally, Rolle’s theorem states that if the outputs of a differentiable function [latex]f[/latex] are equal at the endpoints of an interval, then there must be an interior point [latex]c[/latex] where [latex]f^{\prime}(c)=0[/latex]. Figure 1 illustrates this theorem.
Rolle’s Theorem
Let [latex]f[/latex] be a continuous function over the closed interval [latex][a,b][/latex] and differentiable over the open interval [latex](a,b)[/latex] such that [latex]f(a)=f(b)[/latex]. There then exists at least one [latex]c \in (a,b)[/latex] such that [latex]f^{\prime}(c)=0[/latex].
Proof
Let [latex]k=f(a)=f(b)[/latex]. We consider three cases:
- [latex]f(x)=k[/latex] for all [latex]x \in (a,b)[/latex].
- There exists [latex]x \in (a,b)[/latex] such that [latex]f(x)>k[/latex].
- There exists [latex]x \in (a,b)[/latex] such that [latex]f(x)
Case 1: If [latex]f(x)=k[/latex] for all [latex]x \in (a,b)[/latex], then [latex]f^{\prime}(x)=0[/latex] for all [latex]x \in (a,b)[/latex].
Case 2: Since [latex]f[/latex] is a continuous function over the closed, bounded interval [latex][a,b][/latex], by the extreme value theorem, it has an absolute maximum. Also, since there is a point [latex]x \in (a,b)[/latex] such that [latex]f(x)>k[/latex], the absolute maximum is greater than [latex]k[/latex]. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point [latex]c \in (a,b)[/latex]. Because [latex]f[/latex] has a maximum at an interior point [latex]c[/latex], and [latex]f[/latex] is differentiable at [latex]c[/latex], by Fermat’s theorem, [latex]f^{\prime}(c)=0[/latex].
Case 3: The case when there exists a point [latex]x \in (a,b)[/latex] such that [latex]f(x) [latex]_\blacksquare[/latex] An important point about Rolle’s theorem is that the differentiability of the function [latex]f[/latex] is critical. If [latex]f[/latex] is not differentiable, even at a single point, the result may not hold. For example, the function [latex]f(x)=|x|-1[/latex] is continuous over [latex][-1,1][/latex] and [latex]f(-1)=0=f(1)[/latex], but [latex]f^{\prime}(c) \ne 0[/latex] for any [latex]c \in (-1,1)[/latex] as shown in the following figure. Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points [latex]c[/latex] where [latex]f^{\prime}(c)=0[/latex]. For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values [latex]c[/latex] in the given interval where [latex]f^{\prime}(c)=0[/latex]. Verify that the function [latex]f(x)=2x^2-8x+6[/latex] defined over the interval [latex][1,3][/latex] satisfies the conditions of Rolle’s theorem. Find all points [latex]c[/latex] guaranteed by Rolle’s theorem. Watch the following video to see the worked solution to Example: Using Rolle’s Theorem and the above Try It. Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions [latex]f[/latex]defined on a closed interval [latex][a,b][/latex] with [latex]f(a)=f(b)[/latex]. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (Figure 5). The Mean Value Theorem states that if [latex]f[/latex] is continuous over the closed interval [latex][a,b][/latex] and differentiable over the open interval [latex](a,b)[/latex], then there exists a point [latex]c \in (a,b)[/latex] such that the tangent line to the graph of [latex]f[/latex] at [latex]c[/latex] is parallel to the secant line connecting [latex](a,f(a))[/latex] and [latex](b,f(b))[/latex]. Let [latex]f[/latex] be continuous over the closed interval [latex][a,b][/latex] and differentiable over the open interval [latex](a,b)[/latex]. Then, there exists at least one point [latex]c \in (a,b)[/latex] such that The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting [latex](a,f(a))[/latex] and [latex](b,f(b))[/latex]. Since the slope of that line is and the line passes through the point [latex](a,f(a))[/latex], the equation of that line can be written as Let [latex]g(x)[/latex] denote the vertical difference between the point [latex](x,f(x))[/latex] and the point [latex](x,y)[/latex] on that line. Therefore, Since the graph of [latex]f[/latex] intersects the secant line when [latex]x=a[/latex] and [latex]x=b[/latex], we see that [latex]g(a)=0=g(b)[/latex]. Since [latex]f[/latex] is a differentiable function over [latex](a,b)[/latex], [latex]g[/latex] is also a differentiable function over [latex](a,b)[/latex]. Furthermore, since [latex]f[/latex] is continuous over [latex][a,b][/latex], [latex]g[/latex] is also continuous over [latex][a,b][/latex]. Therefore, [latex]g[/latex] satisfies the criteria of Rolle’s theorem. Consequently, there exists a point [latex]c \in (a,b)[/latex] such that [latex]g^{\prime}(c)=0[/latex]. Since we see that Since [latex]g^{\prime}(c)=0[/latex], we conclude that [latex]_\blacksquare[/latex] In the next example, we show how the Mean Value Theorem can be applied to the function [latex]f(x)=\sqrt{x}[/latex] over the interval [latex][0,9][/latex]. The method is the same for other functions, although sometimes with more interesting consequences. For [latex]f(x)=\sqrt{x}[/latex] over the interval [latex][0,9][/latex], show that [latex]f[/latex] satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value [latex]c \in (0,9)[/latex] such that [latex]f^{\prime}(c)[/latex] is equal to the slope of the line connecting [latex](0,f(0))[/latex] and [latex](9,f(9))[/latex]. Find these values [latex]c[/latex] guaranteed by the Mean Value Theorem. One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 hr down a straight road with an average velocity of 45 mph. Let [latex]s(t)[/latex] and [latex]v(t)[/latex] denote the position and velocity of the car, respectively, for [latex]0 \le t \le 1[/latex] hr. Assuming that the position function [latex]s(t)[/latex] is differentiable, we can apply the Mean Value Theorem to conclude that, at some time [latex]c \in (0,1)[/latex], the speed of the car was exactly If a rock is dropped from a height of 100 ft, its position [latex]t[/latex] seconds after it is dropped until it hits the ground is given by the function [latex]s(t)=-16t^2+100[/latex]. Watch the following video to see the worked solution to Example: Mean Value Theorem and Velocity. Suppose a ball is dropped from a height of 200 ft. Its position at time [latex]t[/latex] is [latex]s(t)=-16t^2+200[/latex]. Find the time [latex]t[/latex] when the instantaneous velocity of the ball equals its average velocity. Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections. At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if [latex]f^{\prime}(x)=0[/latex] for all [latex]x[/latex] in some interval [latex]I[/latex], then [latex]f(x)[/latex] is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly. Let [latex]f[/latex] be differentiable over an interval [latex]I[/latex]. If [latex]f^{\prime}(x)=0[/latex] for all [latex]x \in I[/latex], then [latex]f(x)[/latex] is constant for all [latex]x \in I[/latex]. Since [latex]f[/latex] is differentiable over [latex]I[/latex], [latex]f[/latex] must be continuous over [latex]I[/latex]. Suppose [latex]f(x)[/latex] is not constant for all [latex]x[/latex] in [latex]I[/latex]. Then there exist [latex]a,b \in I[/latex], where [latex]a \ne b[/latex] and [latex]f(a) \ne f(b)[/latex]. Choose the notation so that [latex]a
Since [latex]f[/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c \in (a,b)[/latex] such that Therefore, there exists [latex]c \in I[/latex] such that [latex]f^{\prime}(c) \ne 0[/latex], which contradicts the assumption that [latex]f^{\prime}(x)=0[/latex] for all [latex]x \in I[/latex]. [latex]_\blacksquare[/latex] From the example above, it follows that if two functions have the same derivative, they differ by, at most, a constant. If [latex]f[/latex] and [latex]g[/latex] are differentiable over an interval [latex]I[/latex] and [latex]f^{\prime}(x)=g^{\prime}(x)[/latex] for all [latex]x \in I[/latex], then [latex]f(x)=g(x)+C[/latex] for some constant [latex]C[/latex]. Let [latex]h(x)=f(x)-g(x)[/latex]. Then, [latex]h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=0[/latex] for all [latex]x \in I[/latex]. By Corollary 1, there is a constant [latex]C[/latex] such that [latex]h(x)=C[/latex] for all [latex]x \in I[/latex]. Therefore, [latex]f(x)=g(x)+C[/latex] for all [latex]x \in I[/latex]. [latex]_\blacksquare[/latex] The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function [latex]f[/latex] is increasing over [latex]I[/latex] if [latex]f(x_1) This fact is important because it means that for a given function [latex]f[/latex], if there exists a function [latex]F[/latex] such that [latex]F^{\prime}(x)=f(x)[/latex]; then, the only other functions that have a derivative equal to [latex]f[/latex] are [latex]F(x)+C[/latex] for some constant [latex]C[/latex]. We discuss this result in more detail later in the chapter. Let [latex]f[/latex] be continuous over the closed interval [latex][a,b][/latex] and differentiable over the open interval [latex](a,b)[/latex]. We will prove 1.; the proof of 2. is similar. Suppose [latex]f[/latex] is not an increasing function on [latex]I[/latex]. Then there exist [latex]a[/latex] and [latex]b[/latex] in [latex]I[/latex] such that [latex]a
Since [latex]f(a) \ge f(b)[/latex], we know that [latex]f(b)-f(a) \le 0[/latex]. Also, [latex]a0[/latex]. We conclude that However, [latex]f^{\prime}(x)>0[/latex] for all [latex]x \in I[/latex]. This is a contradiction, and therefore [latex]f[/latex] must be an increasing function over [latex]I[/latex]. [latex]_\blacksquare[/latex]Example: Using Rolle’s Theorem
Try It
The Mean Value Theorem and Its Meaning
Mean Value Theorem
Proof
Example: Verifying that the Mean Value Theorem Applies
Example: Mean Value Theorem and Velocity
Try It
Try It
Corollaries of the Mean Value Theorem
Corollary 1: Functions with a Derivative of Zero
Proof
Corollary 2: Constant Difference Theorem
Proof
Corollary 3: Increasing and Decreasing Functions
Proof
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