4.2: The Mean Value Theorem (2024)

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    Learning Objectives

    • Explain the meaning of Rolle’s theorem.
    • Describe the significance of the Mean Value Theorem.
    • State three important consequences of the Mean Value Theorem.

    The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.

    Rolle’s Theorem

    Informally, Rolle’s theorem states that if the outputs of a differentiable function \(f\) are equal at the endpoints of an interval, then there must be an interior point \(c\) where \(f'(c)=0\). Figure \(\PageIndex{1}\) illustrates this theorem.

    4.2: The Mean Value Theorem (1)

    Rolle’s Theorem

    Let \(f\) be a continuous function over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\) such that \(f(a)=f(b)\). There then exists at least one \(c∈(a,b)\) such that \(f'(c)=0.\)

    Proof

    Let \(k=f(a)=f(b).\) We consider three cases:

    1. \(f(x)=k\) for all \(x∈(a,b).\)
    2. There exists \(x∈(a,b)\) such that \(f(x)>k.\)
    3. There exists \(x∈(a,b)\) such that \(f(x)<k.\)

    Case 1: If \(f(x)=k\) for all \(x∈(a,b)\), then \(f'(x)=0\) for all \(x∈(a,b).\)

    Case 2: Since \(f\) is a continuous function over the closed, bounded interval \([a,b]\), by the extreme value theorem, it has an absolute maximum. Also, since there is a point \(x∈(a,b)\) such that \(f(x)>k\), the absolute maximum is greater than \(k\). Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point \(c∈(a,b)\). Because \(f\) has a maximum at an interior point \(c\), and \(f\) is differentiable at \(c\), by Fermat’s theorem, \(f'(c)=0.\)

    Case 3: The case when there exists a point \(x∈(a,b)\) such that \(f(x)<k\) is analogous to case 2, with maximum replaced by minimum.

    An important point about Rolle’s theorem is that the differentiability of the function \(f\) is critical. If \(f\) is not differentiable, even at a single point, the result may not hold. For example, the function \(f(x)=|x|−1\) is continuous over \([−1,1]\) and \(f(−1)=0=f(1)\), but \(f'(c)≠0\) for any \(c∈(−1,1)\) as shown in the following figure.

    4.2: The Mean Value Theorem (2)

    Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points \(c\) where \(f'(c)=0.\)

    Example \(\PageIndex{1}\): Using Rolle’s Theorem

    For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values \(c\) in the given interval where \(f'(c)=0.\)

    1. \(f(x)=x^2+2x\) over \([−2,0]\)
    2. \(f(x)=x^3−4x\) over \([−2,2]\)

    Solution

    a. Since \(f\) is a polynomial, it is continuous and differentiable everywhere. In addition, \(f(−2)=0=f(0).\) Therefore, \(f\) satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value \(c∈(−2,0)\) such that \(f'(c)=0\). Since \(f'(x)=2x+2=2(x+1),\) we see that \(f'(c)=2(c+1)=0\) implies \(c=−1\) as shown in the following graph.

    4.2: The Mean Value Theorem (3)

    b. As in part a. \(f\) is a polynomial and therefore is continuous and differentiable everywhere. Also, \(f(−2)=0=f(2).\) That said, \(f\) satisfies the criteria of Rolle’s theorem. Differentiating, we find that \(f'(x)=3x^2−4.\) Therefore, \(f'(c)=0\) when \(x=±\frac{2}{\sqrt{3}}\). Both points are in the interval \([−2,2]\), and, therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.

    4.2: The Mean Value Theorem (4)

    Exercise \(\PageIndex{1}\)

    Verify that the function \(f(x)=2x^2−8x+6\) defined over the interval \([1,3]\) satisfies the conditions of Rolle’s theorem. Find all points \(c\) guaranteed by Rolle’s theorem.

    Hint

    Find all values \(c\), where \(f'(c)=0\).

    Answer

    \(c=2\)

    The Mean Value Theorem and Its Meaning

    Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions \(f\) that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (Figure \(\PageIndex{5}\)). The Mean Value Theorem states that if \(f\) is continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\), then there exists a point \(c∈(a,b)\) such that the tangent line to the graph of \(f\) at \(c\) is parallel to the secant line connecting \((a,f(a))\) and \((b,f(b)).\)

    4.2: The Mean Value Theorem (5)

    Mean Value Theorem

    Let \(f\) be continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\). Then, there exists at least one point \(c∈(a,b)\) such that

    \[f'(c)=\frac{f(b)−f(a)}{b−a} \nonumber \]

    Proof

    The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting \((a,f(a))\) and \((b,f(b)).\) Since the slope of that line is

    \[\frac{f(b)−f(a)}{b−a} \nonumber \]

    and the line passes through the point \((a,f(a)),\) the equation of that line can be written as

    \[y=\frac{f(b)−f(a)}{b−a}(x−a)+f(a). \nonumber \]

    Let \(g(x)\) denote the vertical difference between the point \((x,f(x))\) and the point \((x,y)\) on that line. Therefore,

    \[g(x)=f(x)−\left[\frac{f(b)−f(a)}{b−a}(x−a)+f(a)\right]. \nonumber \]

    Since the graph of \(f\) intersects the secant line when \(x=a\) and \(x=b\), we see that \(g(a)=0=g(b)\). Since \(f\) is a differentiable function over \((a,b)\), \(g\) is also a differentiable function over \((a,b)\). Furthermore, since \(f\) is continuous over \([a,b], \, g\) is also continuous over \([a,b]\). Therefore, \(g\) satisfies the criteria of Rolle’s theorem. Consequently, there exists a point \(c∈(a,b)\) such that \(g'(c)=0.\) Since

    \[g'(x)=f'(x)−\frac{f(b)−f(a)}{b−a}, \nonumber \]

    we see that

    \[g'(c)=f'(c)−\frac{f(b)−f(a)}{b−a}. \nonumber \]

    Since \(g'(c)=0,\) we conclude that

    \[f'(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \]

    In the next example, we show how the Mean Value Theorem can be applied to the function \(f(x)=\sqrt{x}\) over the interval \([0,9]\). The method is the same for other functions, although sometimes with more interesting consequences.

    Example \(\PageIndex{2}\): Verifying that the Mean Value Theorem Applies

    For \(f(x)=\sqrt{x}\) over the interval \([0,9]\), show that \(f\) satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value \(c∈(0,9)\) such that \(f′(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9))\). Find these values \(c\) guaranteed by the Mean Value Theorem.

    Solution

    We know that \(f(x)=\sqrt{x}\) is continuous over \([0,9]\) and differentiable over \((0,9).\) Therefore, \(f\) satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value \(c∈(0,9)\) such that \(f′(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9))\) (Figure \(\PageIndex{7}\)). To determine which value(s) of \(c\) are guaranteed, first calculate the derivative of \(f\). The derivative \(f′(x)=\frac{1}{(2\sqrt{x})}\). The slope of the line connecting \((0,f(0))\) and \((9,f(9))\) is given by

    \[\frac{f(9)−f(0)}{9−0}=\frac{\sqrt{9}−\sqrt{0}}{9−0}=\frac{3}{9}=\frac{1}{3}. \nonumber \]

    We want to find \(c\) such that \(f′(c)=\frac{1}{3}\). That is, we want to find \(c\) such that

    \[\frac{1}{2\sqrt{c}}=\frac{1}{3}. \nonumber \]

    Solving this equation for \(c\), we obtain \(c=\frac{9}{4}\). At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

    4.2: The Mean Value Theorem (7)

    One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let \(s(t)\) and \(v(t)\) denote the position and velocity of the car, respectively, for \(0≤t≤1\) h. Assuming that the position function \(s(t)\) is differentiable, we can apply the Mean Value Theorem to conclude that, at some time \(c∈(0,1)\), the speed of the car was exactly

    \[v(c)=s′(c)=\frac{s(1)−s(0)}{1−0}=45\,\text{mph.} \nonumber \]

    Example \(\PageIndex{3}\): Mean Value Theorem and Velocity

    If a rock is dropped from a height of 100 ft, its position \(t\) seconds after it is dropped until it hits the ground is given by the function \(s(t)=−16t^2+100.\)

    1. Determine how long it takes before the rock hits the ground.
    2. Find the average velocity \(v_{avg}\) of the rock for when the rock is released and the rock hits the ground.
    3. Find the time \(t\) guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is \(v_{avg}.\)

    Solution

    a. When the rock hits the ground, its position is \(s(t)=0\). Solving the equation \(−16t^2+100=0\) for \(t\), we find that \(t=±\frac{5}{2}sec\). Since we are only considering \(t≥0\), the ball will hit the ground \(\frac{5}{2}\) sec after it is dropped.

    b. The average velocity is given by

    \[v_{avg}=\frac{s(5/2)−s(0)}{5/2−0}=\frac{0−100}{5/2}=−40\,\text{ft/sec}. \nonumber \]

    c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time \(t\) such that \(v(t)=s′(t)=v_{avg}=−40\) ft/sec. Since \(s(t)\) is continuous over the interval \([0,5/2]\) and differentiable over the interval \((0,5/2),\) by the Mean Value Theorem, there is guaranteed to be a point \(c∈(0,5/2)\) such that

    \[s′(c)=\frac{s(5/2)−s(0)}{5/2−0}=−40. \nonumber \]

    Taking the derivative of the position function \(s(t)\), we find that \(s′(t)=−32t.\) Therefore, the equation reduces to \(s′(c)=−32c=−40.\) Solving this equation for \(c\), we have \(c=\frac{5}{4}\). Therefore, \(\frac{5}{4}\) sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: \(−40\) ft/sec.

    4.2: The Mean Value Theorem (8)

    Exercise \(\PageIndex{2}\)

    Suppose a ball is dropped from a height of 200 ft. Its position at time \(t\) is \(s(t)=−16t^2+200.\) Find the time \(t\) when the instantaneous velocity of the ball equals its average velocity.

    Hint

    First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground.

    Answer

    \(\frac{5}{2\sqrt{2}}\) sec

    Corollaries of the Mean Value Theorem

    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 \(f′(x)=0\) for all \(x\) in some interval \(I\), then \(f(x)\) 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.

    Corollary 1: Functions with a Derivative of Zero

    Let \(f\) be differentiable over an interval \(I\). If \(f′(x)=0\) for all \(x∈I\), then \(f(x)=\) constant for all \(x∈I.\)

    Proof

    Since \(f\) is differentiable over \(I\), \(f\) must be continuous over \(I\). Suppose \(f(x)\) is not constant for all \(x\) in \(I\). Then there exist \(a,b∈I\), where \(a≠b\) and \(f(a)≠f(b).\) Choose the notation so that \(a<b.\) Therefore,

    \[\frac{f(b)−f(a)}{b−a}≠0. \nonumber \]

    Since \(f\) is a differentiable function, by the Mean Value Theorem, there exists \(c∈(a,b)\) such that

    \[f′(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \]

    Therefore, there exists \(c∈I\) such that \(f′(c)≠0\), which contradicts the assumption that \(f′(x)=0\) for all \(x∈I\).

    From "Corollary 1: Functions with a Derivative of Zero,"it follows that if two functions have the same derivative, they differ by, at most, a constant.

    Corollary 2: Constant Difference Theorem

    If \(f\) and \(g\) are differentiable over an interval \(I\) and \(f′(x)=g′(x)\) for all \(x∈I\), then \(f(x)=g(x)+C\) for some constant \(C\).

    Proof

    Let \(h(x)=f(x)−g(x).\) Then, \(h′(x)=f′(x)−g′(x)=0\) for all \(x∈I.\) By Corollary 1, there is a constant \(C\) such that \(h(x)=C\) for all \(x∈I\). Therefore, \(f(x)=g(x)+C\) for all \(x∈I.\)

    The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function \(f\) is increasing over \(I\) if \(f(x_1)<f(x_2)\) whenever \(x_1<x_2\), whereas \(f\) is decreasing over \(I\) if \(f(x_1)>f(x_2)\) whenever \(x_1<x_2\). Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure \(\PageIndex{9}\)). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

    This fact is important because it means that for a given function \(f\), if there exists a function \(F\) such that \(F′(x)=f(x)\); then, the only other functions that have a derivative equal to \(f\) are \(F(x)+C\) for some constant \(C\). We discuss this result in more detail later in the chapter.

    4.2: The Mean Value Theorem (9)

    Corollary 3: Increasing and Decreasing Functions

    Let \(f\) be continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\).

    1. If \(f′(x)>0\) for all \(x∈(a,b)\), then \(f\) is an increasing function over \([a,b].\)
    2. If \(f′(x)<0\) for all \(x∈(a,b)\), then \(f\) is a decreasing function over \([a,b].\)

    Proof

    We will prove i.; the proof of ii. is similar. Suppose \(f\) is not an increasing function on \(I\). Then there exist \(a\) and \(b\) in \(I\) such that \(a<b\), but \(f(a)≥f(b)\). Since \(f\) is a differentiable function over \(I\), by the Mean Value Theorem there exists \(c∈(a,b)\) such that

    \[f′(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \]

    Since \(f(a)≥f(b)\), we know that \(f(b)−f(a)≤0\). Also, \(a<b\) tells us that \(b−a>0.\) We conclude that

    \[f′(c)=\frac{f(b)−f(a)}{b−a}≤0. \nonumber \]

    However, \(f′(x)>0\) for all \(x∈I\). This is a contradiction, and therefore \(f\) must be an increasing function over \(I\).

    Key Concepts

    • If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\) and \(f(a)=f(b)\), then there exists a point \(c∈(a,b)\) such that \(f′(c)=0.\) This is Rolle’s theorem.
    • If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\), then there exists a point \(c∈(a,b)\) such that \[f'(c)=\frac{f(b)−f(a)}{b−a}. \nonumber \] This is the Mean Value Theorem.
    • If \(f'(x)=0\) over an interval \(I\), then \(f\) is constant over \(I\).
    • If two differentiable functions \(f\) and \(g\) satisfy \(f′(x)=g′(x)\) over \(I\), then \(f(x)=g(x)+C\) for some constant \(C\).
    • If \(f′(x)>0\) over an interval \(I\), then \(f\) is increasing over \(I\). If \(f′(x)<0\) over \(I\), then \(f\) is decreasing over \(I\).

    Glossary

    mean value theorem

    if \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\), then there exists \(c∈(a,b)\) such that \(f′(c)=\frac{f(b)−f(a)}{b−a}\)

    rolle’s theorem
    if \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\), and if \(f(a)=f(b)\), then there exists \(c∈(a,b)\) such that \(f′(c)=0\)
    4.2: The Mean Value Theorem (2024)

    FAQs

    What is the 4.2 Mean Value Theorem? ›

    Figure 4.2. 5: The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values c1 and c2 such that the tangent line to f at c1 and c2 has the same slope as the secant line.

    What is the answer to the Mean Value Theorem? ›

    The Mean Value Theorem states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b].

    What is the Mean Value Theorem quizlet? ›

    mean value theorem. if f(x) is defined on [a,b], differentiable on (a,b) and f(a)=f(b) then there is a number c between and b and f'(c)=0.

    How to determine if the Mean Value Theorem applies? ›

    "The precise conditions under which MVT applies are that fff is differentiable over the open interval (a,b) and continuous over the closed interval [a,b]. Since differentiability implies continuity, we can also describe the condition as being differentiable over (a,b) and continuous at x=a and x=b."

    What is the 4 number theorem? ›

    Four Number Theorem: any two factorizations ab=cd have a common refinement. Let a, b, c, d be integers such that ab=cd. Then there exist integers x, y, z, w such that xy=a, zw=b, xz=c, yw=d.

    What is theorem 4 4? ›

    Theorem 4-4 If two angles are complementary to the same angle, then they are congruent to each other. Theorem 4-5 If two angles are complementary to two congruent angles, then the two angles are congruent to each. other.

    How to prove mean value theorem? ›

    Mean Value Theorem Proof

    h(a) = h(b) = 0 and h(x) is continuous on [a, b] and differentiable on (a, b). Thus applying the Rolles theorem, there is some x = c in (a, b) such that h'(c) = 0. Thus the mean value theorem is proved.

    What is the result of the mean value theorem? ›

    Mean Value Theorem states that for any function f(x) passing through two given points [a, f(a)], [b, f(b)], there exists at least one point [c, f(c)] on the curve such that the tangent through that point is parallel to the secant passing through the other two points.

    What is the mean value theorem conclusion? ›

    The conclusion is that there exists a point in the interval such that the tangent at the point c , f c is parallel to the line that passes through the points a , f a and b , f b .

    What is the mean value theorem made easy? ›

    The derivative at a point is the same thing as the slope of the tangent line at that point, so the theorem just says that there must be at least one point between a and b where the slope of the tangent line is the same as the slope of the secant line from a to b.

    What is the mean value theorem measure? ›

    The Essentials. The Mean Value Theorem states that if a function f is continuous over [a,b] and differentiable over (a,b), then at some point, c, along the function, the average slope of f over [a,b] is equal to the instantaneous slope at f(c). f ′ c = f b - f a b - a.

    What is the mean value theorem argument? ›

    In mathematics, the mean value theorem (or Lagrange theorem) states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. It is one of the most important results in real analysis.

    What does the Mean Value Theorem help you find? ›

    The mean value theorem connects the average rate of change of a function to its derivative. It says that for any differentiable function ‍ and an interval ‍ (within the domain of ‍ ), there exists a number ‍ within ‍ such that ‍ is equal to the function's average rate of change over ‍ .

    When would you use Mean Value Theorem? ›

    So the Mean Value Theorem (MVT) allows us to determine a point within the interval where both the slope of the tangent and secant lines are equal. Now, let's think geometrically for a second. If two linear are parallel, then we know that they have the same slope. This means we are on the hunt for parallel lines.

    Why does the Mean Value Theorem matter? ›

    It can be used to establish results and estimates. For example that f'=g' implies f=g+c, the fundamental theorem of calculus, Taylor's theorem with remainder, or the error estimates for Simpson's rule.

    What is the theorem 2.1 4? ›

    4: Dedekind Theorem. A set S is infinite if and only if there exists a proper subset A of S which has the same cardinality as S.

    What is the theorem 1.7 4? ›

    Theorem 1.7. 3: If two angles are supplementary to the same angle (or to congruent angles, then the angles are congruent. Theorem 1.7. 4: Any two right angles are congruent.

    What is the Mean Value Theorem expression? ›

    f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange's mean value theorem.

    What is the Mean Value Theorem measure? ›

    The Essentials. The Mean Value Theorem states that if a function f is continuous over [a,b] and differentiable over (a,b), then at some point, c, along the function, the average slope of f over [a,b] is equal to the instantaneous slope at f(c). f ′ c = f b - f a b - a.

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