The Mean Value Theorem A student at the University of Connecticut - - PDF document

The Mean Value Theorem

A student at the University of Connecticut happens to be travelling to Boston. He enters the Massachussetts Turnpike at the Sturbridge Village entrance at 9:15 in the morning. Since he uses Fast Lane, he never actually picksup a ticket, but sensorsrecord that hegoesthrough the toll lane at Newton at 9:53, just 38 minutes later. He soon receives a traffic summons in the mail indicating that he vi-

• lated the posted speed limit and decides to appeal, since there is no

indication he was caught by radar. Appearing in court, theprosecutor arguesthat hetravelled 44.8 miles in 38 minutesand thereforetravelled at an averagespeed of 70.74 miles per hour, more than 5 miles above the highest posted speed limit of 65 miles per hour. The student argues that, although he may have averaged more than 65 miles per hour, there is no evidence that he actually ever travelled at an instantaneous speed of more than 65 miles per hour. The judge considers the arguments and quickly rejects the student’s plea, noting that the Mean Value Theorem implies that since his aver- agespeed was 70.74 miles per hour, therehad to beat least oneinstant during which his instantaneous speed was 70.74 miles per hour.

Rolle’s Theorem

Theorem 1 (Rolle’s Theorem). Suppose a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b) and f(a) = f(b). Then there is a number c ∈ (a, b) such that f ′(c) = 0. Geometrically, this says that if there are two points on a smooth curve takes at the same height, there must be a point in between where the tangent is horizontal. Rolle’s Theorem is a special case of the Mean Value Theorem. Proof: Suppose f satisfies the hypotheses of Rolle’s Theorem. By the Extreme Value Theorem for Continuous Function, there must be some point in [a, b] at which f attains a minimum and some point at which f attains a maximum. One possibility is that f is constant on the entire interval, in which case f ′ is identically 0 on (a, b) and the conclusion is clearly true. So let’s consider the other possibility, that f is not constant. Then either the minimum or the maximum must occur at some point c ∈ (a, b), that is, at some point other than the endpoints.

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Wewill show that f ′(c) = 0 if f hasa maximum at c. Similar reasoning would show f ′(c) = 0 if f had a minimum at c, showing the conclusion is true and completing the proof. We know f ′(c) = limx→c f(x) − f(c) x − c . Since this ordinary limit exists, it follows that both the left hand limit and the right hand limit exist and are both equal to f ′(c). We’ll con- sider them separately. The Left Hand Limit: f ′(c) = limx→c− f(x) − f(c) x − c . Sincef hasa maximum at c, if x < c, then f(x) ≤ f(c), so f(x)−f(c) ≤

• 0. But, if x < c, it’s also true that x − c < 0 and it follows that

f(x) − f(c) x − c ≥ 0. We see limx→c− f(x) − f(c) x − c is the limit of non- negative numbers and therefore can’t be negative. It follows that f ′(c) ≥ 0. The line of reasoning for the right hand limit is similar. The Right Hand Limit: f ′(c) = limx→c+ f(x) − f(c) x − c . Sincef hasa maximum at c, if x > c, then f(x) ≤ f(c), so f(x)−f(c) ≤

• 0. But, if x > c, it’s also true that x − c > 0 and it follows that

f(x) − f(c) x − c ≤ 0. We see limx→c+ f(x) − f(c) x − c is the limit of numbers less than or equal to 0 and therefore can’t be positive. It follows that f ′(c) ≤ 0. Since f ′(c) ≥ 0 and f ′(c) ≤ 0, it follows that f ′(c) = 0 QED

Consequences of Rolle’s Theorem

Besides being a special case of the Mean Value Theorem and being a step in the path to proving the Mean Value Theorem, Rolle’s Theorem has some interesting applications of its own. Corollary 2. A polynomial equation of degree n has at most n solu- tions. This is equivalent to the statement that a polynomial of degree n has at most n zeros. Sincea polynomial may bedifferentiated asmany timesasnecessary, with each derivativebeing a polynomial of lower degree, oneimmediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial.

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The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. Thederivativeof a quadratic polynomial islinear, having no morethan 1 zero, so a quadratic can’t have more than 2 zeros. The derivative of a cubic polynomial is quadratic, having no morethan 2 zeros, so the cubic can’t have more than 3 zeros. This clearly goes on forever. The argument can be made rigorous through the use of Mathematical Induction.

The Mean Value Theorem

Theorem 3 (The Mean Value Theorem). Suppose a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there is a number c ∈ (a, b) such that f(b)−f(a) = f ′(c)(b − a) or, equivalently, f ′(c) = f(b) − f(a) b − a . Geometrically, the Mean Value Theorem says that if there is a smooth curve connecting two points, there must be some point in between at which the tangent line is parallel to the line connecting those two points. Analytically, the Mean Value Theorem says the rate of change of a differentiable function must, at some point, take on its average, or mean value. The proof of the Mean Value Theorem depends on the fact that the particular point at which the tangent line is parallel to the line connecting the endpoints also happens to be the point at which the curve is furthers away from that line. The proof essentially consists

• f applying Rolle’s Theorem to the function measuring the distance

between the line and the curve. Since the line goes between the points (a, f(a)) and (b, f(b)), its slope will be f(b) − f(a) b − a and its equation may be written, in point- slope form, y − f(a) = f(b) − f(a) b − a · (x − a). Solving for y, we may write the equation in the form y = f(a) + f(b) − f(a) b − a · (x − a). Sincethesecond coordinateof a point on thecurvewith first coordinate x isf(x), thevertical distancebetween thelineand thecurvewill equal

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f(x) −

• f(a) + f(b) − f(a)

b − a · (x − a)

• =

f(x) − f(a) − f(b) − f(a) b − a · (x − a). We are now prepared to prove the Mean Value Theorem. Proof: Let φ(x) = f(x) − f(a) − f(b) − f(a) b − a · (x − a). It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [a, b]. Certainly, the fact that φ is both continuous and differentiable on [a, b] follows immediately from the fact that f is. In addition, φ(a) = f(a) − f(a) − f(b) − f(a) b − a · (a − a) = 0 and φ(b) = f(b)−f(a)−f(b) − f(a) b − a ·(b−a) = f(b)−f(a)−[f(b)−f(a)] = 0. Thus, there must be some c ∈ (a, b) such that φ′(c) = 0. We first

• btain φ′(c) as follows.

φ′(x) = f ′(x) − f(b) − f(a) b − a φ′(c) = f ′(c) − f(b) − f(a) b − a Since φ′(c) = 0, it follows that f ′(c) − f(b) − f(a) b − a = 0. f ′(c) = f(b) − f(a) b − a f ′(c)(b − a) = f(b) − f(a) QED

Consequences of the Mean Value Theorem

Perhapsthemost important consequenceof theMean Value Theorem is that it gives precise meaning to the most important single concept in elementary Calculus,

The Derivative Measures Rate of Change.

Theorem 4.

• a. If the derivative of a function is positive at all points on an

interval, then the function is increasing on that interval.

• b. If the derivative of a function is negative at all points on an

interval, then the function is decreasing on that interval.

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To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition 1 (Strictly Increasing). A function f is said to be strictly increasing on an open interval I if f(a) < f(b) whenever a, b ∈ I and a < b. Definition 2 (Nondecreasing). A function f is said to be nondecreas- ing on an open interval I if f(a) ≤ f(b) whenever a, b ∈ I and a < b. Note the subtle difference. Often, we will simply say a function is

• increasing. Generally, in those cases, it will not really be important

whether we mean strictly increasing or simply nondecreasing. There are similar definitions of the terms strictly decreasing and non-

• increasing. Here, too, we will often use the ambiguous term decreasing.

Definition 3 (Strictly Decreasing). A function f is said to be strictly decreasing on an open interval I if f(a) > f(b) whenever a, b ∈ I and a < b. Definition 4 (Nonincreasing). A function f is said to be nonincreasing

• n an open interval I if f(a) ≥ f(b) whenever a, b ∈ I and a < b.

With these definitions, we are ready to prove the derivative measures rate of change. We will prove just one of what are really four different parts, that if the derivative is strictly positive then the function is strictly increasing.

• Proof. Suppose f ′(x) > 0

∀x ∈ I and let a, b ∈ I, a < b. We need to prove that f(a) < f(b). Note: We’ve introduced the notation ∀ to mean for all or for every. Since f ′(x) > 0 ∀x ∈ I, it follows that f is continuous on [a, b] and differentiable on (a, b), so by the Mean Value Theorem there is some c ∈ (a, b) such that f(b) − f(a) = f ′(c)(b − a). Since f ′(x) > 0 ∀x ∈ I, it follows that f ′(c) is positive. Since a < b, it follows that b − a is also positive, so that f ′(c)(b − a) is also positive and hence f(b) − f(a) must be positive. Clearly, f(a) must be smaller than f(b).

• Very Different Functions Can’t Have the Same Derivative

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It’s obvious that if two functions differ by only a constant term, then they will have the same derivative. Example: d dx (x2) = d dx (x2 + 5) = 2x. A natural question is whether only functions differing by a constant can share the same derivative. The Mean Value Theorem enables us to see this is true. Theorem 5. If f ′(x) = g′(x) for all x in some interval, then there is some constant k such that f(x) = g(x) + k. Proof: Let φ = f − g and let a be some fixed point in the interval. Now let x be in the interval. Clearly, the φ is both continuous and differentiable on the interval with endpoints a and x and we can apply the Mean Value Theorem. We use that language because it is possible that a < x and the interval is [a, x] but also possible that a > x and the interval is [x, a]. By the Mean Value Theorem, there is some c in the interval such that φ(x) − φ(a) = φ′(c)(x − a). Because φ = f − g, it follows that φ′(c) = 0, so φ(x) − φ(a) = 0. It follows that φ(x) = φ(a), or φ(x) = k, where we let k = φ(a). Since φ = f − g, we have f(x) − g(x) = k, or f(x) = g(x) + k. QED Corollary 6. Only constant functions have derivatives which are iden- tically 0. This theorem will prove useful when we try to find functions with a given derivative. One application will be to determine the acceleration due to gravity.

Example

An object is dropped from a height of 128 feet. How long does it take to reach the ground?

Solution

Let:

• h be the height of the object, measured in feet.
• v be the velocity of the object, measured in feet per second.
• t bethetimesincetheobject wasdropped, measured in seconds.

We Know:

• h = 128 when t = 0
• v = 0 when t = 0

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• dh

dt = v (Since velocity is the rate at which the height changes.)

• dv

dt = −32 (Since acceleration is the rate at which the velocity changesand theacceleration dueto gravity is32 feet per second per second and is in the downward, or negative, direction.) frame

We Want:

• The value of t when h = 0

Since d dt (−32t) = −32 and two functions can have the same deriva- tive only if they differ by a constant, it follows that v = −32t + c for some constant c. Since v = 0 when t = 0, it follows that 0 = −32 · 0 + c c = 0 v = −32t Since d dt (−16t2) = −32t, dh dt = v = −32t and two functions can have the same derivative only if they differ by a constant, it follows that h = −16t2 + k for some constant k. Since h = 128 when t = 0, it follows that 128 = −16 · 02 + k k = 128 h = −16t2 + 128 Thus, when h = 0, 0 = −16t2 + 128 16t2 = 128 t2 = 128 16 = 8 t = √ 8 = 2 √ 2 ≈ 2.828 It thus takes 2 √ 2 seconds for the object to fall to the ground. We can also figure out how fast it was going when it hit the ground: Since v = −32t, when t = 2 √ 2 we have v = −32 · 2 √ 2 = −64 √ 2 ≈ −90.510, so the object is travelling at a speed of 64 √ 2 feet per second when it hits the ground.

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Example

Determine the acceleration due to gravity.

Solution:

We may perform the following experiment: We drop a coin from the ceiling and measure the time it takes to hit the ground. Let:

• t be the time since the coin is dropped
• T be the amount of time it takes to hit the ground
• h be the height of the coin
• H be the height of the ceiling
• v be the speed of the coin
• g be the acceleration due to gravity

We Know

We know:

• dh

dt = v

• dv

dt = g

• h = H when t = 0
• v = 0 when t = 0
• h = 0 when t = T

We Want

We want:

• The value of g.

Since d dt (gt) = g, dv dt = g and two functions can have the same derivative only if they differ by a constant, it follows that v = gt + c for some constant c. Since v = 0 when t = 0, it follows that 0 = g · 0 + c c = 0 v = gt Since d dt 1

2gt2

= gt, dh dt = v = gt and two functions can have the same derivative only if they differ by a constant, it follows that h =

1 2gt2 + k for some constant k.

Since h = H when t = 0, it follows that

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H =

1 2g · 02 + k

k = H h =

1 2gt2 + H

Since h = 0 when t = T, it follows that 0 =

1 2gT 2 + H 1 2gT 2 = −H

g = −H

1 2T 2

g = −2H T 2 So the acceleration due to gravity is 2H T 2 feet per second per second. Note that g is negative, since the acceleration is downward.