3.2: Mean Value Theorem

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If you averaged 30 miles per hour during a trip, then at some instant during the trip you were traveling exactly 30 miles per hour.

That relatively obvious statement is the Mean Value Theorem as it applies to a particular trip. It may seem strange that such a simple statement would be important or useful to anyone, but the Mean Value Theorem is important and some of its consequences are very useful in a variety of areas. Many of the results in the rest of this chapter depend on the Mean Value Theorem, and one of the corollaries of the Mean Value Theorem will be used every time we calculate an “integral” in later chapters. A truly delightful aspect of mathematics is that an idea as simple and obvious as the Mean Value Theorem can be so powerful.

Before we state and prove the Mean Value Theorem and examine some of its consequences, we will consider a simplified version called Rolle’s Theorem.

Rolle’s Theorem

Pick any two points on the $x$-axis and think about all of the differentiable functions that pass through those two points.

$This illustration shows four distinct curves (red, blue, magenta and brown) plotted on the same coordinate plane, all starting and ending at the same two points on the horizontal axis. The red curve follows a complex path with three local peaks and two local valleys. The blue curve forms a single, tall smooth arc. The magenta curve stays closer to the horizontal axis with two gentle rounded hills and one shallow valley. Finally, the brown curve is reflected across the horizontal axis, staying entirely below it while forming two rounded valleys.$

Because our functions are differentiable, they must be continuous and their graphs cannot have any holes or breaks. Also, since these functions are differentiable, their derivatives are defined everywhere between our two points and their graphs can not have any “corners” or vertical tangents. The graphs of the functions in the margin figure can still have all sorts of shapes, and it may seem unlikely that they have any common properties other than the ones we have stated, but Michel Rolle (1652–1719) found one. He noticed that every one of these functions has one or more points where the tangent line is horizontal:

and this result is named after him.

Rolle’s Theorem

If: $f(a) = f(b)$

and: $f(x)$ is continuous for $a \leq x \leq b$ and differentiable for $a < x < b$

then: there is at least one number $c$ between $a$ and $b$ so that $f'(c) = 0$.

Proof

We consider three cases: when $f(x) = f(a)$ for all $x$ in $(a,b)$, when $f(x) > f(a)$ for some $x$ in $(a,b)$, and when $f(x) < f(a)$ for some $x$ in $(a,b)$.

Case I: If $f(x) = f(a)$ for all $x$ between $a$ and $b$, then the graph of $f$ is a horizontal line segment and $f'(c) = 0$ for all values of $c$ strictly between $a$ and $b$.

Case II: Suppose $f(x) > f(a)$ for some $x$ in $(a,b)$. Because $f$ is continuous on the closed interval $[a,b]$, we know from the Extreme Value Theorem that $f$ must attain a maximum value on the closed interval $[a,b]$. Because $f(x) > f(a)$ for some value of $x$ in $[a,b]$, then the maximum of $f$ must occur at some value $c$ strictly between $a$ and $b$: $a < c < b$. (Why can't the maximum be at $a$ or $b$?) Because $f(c)$ is a local maximum of $f$, $c$ is a critical number of $f$, meaning $f'(c) = 0$ or $f'(c)$ is undefined. But $f$ is differentiable at all $x$ between $a$ and $b$, so the only possibility is that $f'(c) = 0$. (Notice that Rolle’s Theorem tells us that [at least one] number $c$ with the required properties exists, but does not tell us how to find $c$.)

Case III: Suppose $f(x) < f(a)$ for some $x$ in $(a,b)$. Then, arguing as we did in Case II, $f$ attains a minimum at some value $x = c$ strictly between $a$ and $b$, and so $f'(c) = 0$.

In each case, there is at least one value of $c$ between $a$ and $b$ so that $f'(c) = 0$.

Example $\PageIndex{1}$

Show that $f(x) = x^3 - 6x^2 + 9x + 2$ satisfies the hypotheses of Rolle’s Theorem on the interval $[0,3]$ and find a value of $c$ that the theorem tells you must exist.

Solution

Because $f$ is a polynomial, it is continuous and differentiable everywhere. Furthermore, $f(0) = 2 = f(3)$, so Rolle’s Theorem applies. Differentiating:\[f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)\nonumber\]so $f'(x) = 0$ when $x = 1$ and when $x = 3$. The value $c = 1$ is between $0$ and $3$. The figure below:

$A red graph of the curve f(x) = x^3 - 6x^2 + 9x + 2 on a [0,3]X[0,6] grid with a back horizontal line segment tangent to the curve at x =1.$

shows a graph of $f$.

Practice $\PageIndex{1}$

Find the value(s) of $c$ for Rolle’s Theorem for the functions graphed below.

$This image displays two side-by-side graphs on coordinate planes, labeled y equals f(x) in red and y equals g(x) in blue, illustrating different rates of change and extrema behavior. Both horizontal axes are marked from zero to seven. The red graph for f(x) shows a very gentle, smooth curve that rises to a broad local maximum near x equals 2 and then gradually slopes downward to a shallow local minimum near x equals 6 before slightly turning upward again. The blue graph for g(x) shows much more dramatic fluctuations. It starts higher on the vertical axis, plunges quickly to a deep local minimum at x equals 2, and then shoots upward to a tall, sharp local maximum near x equals 4. After this peak, it drops into another local minimum near x equals 6 before rising at the end.$

Answer

$f'(x) = 0$ when $x = 2$ and $6$, so $c = 2$ and $c= 6$.

$g'(x) = 0$ when $x = 2$, $4$ and $6$, so $c = 2$, $c = 4$ and $c = 6$.

The Mean Value Theorem

Geometrically, the Mean Value Theorem is a “tilted” version of Rolle’s Theorem:

$A red curve (labeled f(x)) and a blue curve (labeled g(x)) start and end at the same two black dots in the first quadrant. The first black dot is at x = a and the second at x = b. A black line segment passes through the two dots. The red curve rises above the line segment and peaks at a point between x = a and x = b. At a point labeled 'c for f(x)' another black line segment is tangent to the red curve. The blue curve rises but stays below the secant segment. At a point labeled 'c for g(x)' a third black line segment is tangent to the blue curve.$

In each theorem we conclude that there is a number $c$ so that the slope of the tangent line to $f$ at $x = c$ is the same as the slope of the line connecting the two ends of the graph of $f$ on the interval $[a,b]$. In Rolle’s Theorem, the two ends of the graph of $f$ are at the same height, $f(a) = f(b)$, so the slope of the line connecting the ends is zero. In the Mean Value Theorem, the two ends of the graph of $f$ do not have to be at the same height, so the line through the two ends does not have to have a slope of zero.

Mean Value Theorem

If: $f(x)$ is continuous for $a \leq x \leq b$ and differentiable for $a < x < b$

then: there is at least one number $c$ between $a$ and $b$ so the line tangent to the graph of $f$ at $x=c$ is parallel to the secant line through $(a, f(a))$ and $(b,f(b))$:\[f'(c) = \frac{f(b) - f(a)}{b - a}\nonumber\]

Proof

The proof of the Mean Value Theorem uses a tactic common in mathematics: introduce a new function that satisfies the hypotheses of some theorem we already know and then use the conclusion of that previously proven theorem. For the Mean Value Theorem we introduce a new function, $h(x)$, which satisfies the hypotheses of Rolle’s Theorem. Then we can be certain that the conclusion of Rolle’s Theorem is true for $h(x)$ and the Mean Value Theorem for $f$ will follow from the conclusion of Rolle’s Theorem for $h$.

First, let $g(x)$ be the linear function passing through the points $(a, f(a))$ and $(b, f(b))$ of the graph of $f$. The function $g$ goes through the point $(a, f(a))$ so $g(a) = f(a)$. Similarly, $g(b) = f(b)$. The slope of the linear function $g(x)$ is $\displaystyle \frac{f(b) - f(a)}{b - a}$ so $\displaystyle g'(x) = \frac{f(b) - f(a)}{b - a}$ for all $x$ between $a$ and $b$, and $g$ is continuous and differentiable. (The formula for $g$ is $g(x) = f(a) + m(x - a)$ with $m = \frac{f(b) - f(a)}{b - a}$.)

Define $h(x) = f(x) - g(x)$ for $a \leq x \leq b$:

$A red curve (labeled f(x)) starts and ends at two black dots in the first quadrant. The first black dot is at x = a and the second at x = b. A blue line segment (labeled g(x)) passes through the two dots. The red curve rises above the line segment and peaks at a point between x = a and x = b. A dashed-black vertical line segment connects a third black dot on the blue segment with the horizontal axis at a point labeled 'x.' A double arrow labeled h(x) connects the third black dot with a fourth black dot immediately above it on the red curve.$

The function $h$ satisfies the hypotheses of Rolle’s theorem:

$h(a) = f(a) - g(a) = 0$ and $h(b) = f(b) - g(b) = 0$
$h(x)$ is continuous for $a \leq x \leq b$ because both $f$ and $g$ are continuous there
$h(x)$ is differentiable for $a < x < b$ because both $f$ and $g$ are differentiable there

so the conclusion of Rolle’s Theorem applies to $h$: there is a $c$ between $a$ and $b$ so that $h'(c) = 0$.

The derivative of $h(x) = f(x) - g(x)$ is $h'(x) = f'(x) - g'(x)$ so we know that there is a number $c$ between $a$ and $b$ with $h'(c) = 0$. But:\[ 0 = h'(c) = f'(c) - g'(c) \Rightarrow f'(c) = g '(c) = \frac{f(b) - f(a)}{b - a}\nonumber\]which is exactly what we needed to prove.

Graphically, the Mean Value Theorem says that there is at least one point $c$ where the slope of the tangent line, $f'(c)$, equals the slope of the line through the end points of the graph segment, $(a, f(a) )$ and $(b, f(b) )$. The figure below shows the locations of the parallel tangent lines for several functions and intervals:

$Each graph in this set of three graphs shows a red function on a coordinate plane, with a dashed secant line connecting two endpoints and solid tangent lines drawn at specific interior points. In the first panel on the left, a dashed secant line slopes downward between two endpoints of a curved function. At two distinct interior locations, labeled with the letter c, the function has tangent lines that are exactly parallel to that dashed secant line. This demonstrates that there can be multiple points where the instantaneous rate of change matches the average rate of change. The middle panel shows a steeply rising function that eventually curves back down. A dashed secant line connects a very low starting point to a higher ending point. There is one interior point, marked with a vertical arrow at location c, where a tangent line sits on top of a rounded peak, perfectly parallel to the slope of the dashed line. The third panel on the right depicts a more complex wavy function with a dashed secant line sloping downward across the entire interval. In this case, three separate interior points are identified with the letter c. At each of these three locations, the solid tangent lines have the same downward tilt as the dashed secant line.$

The Mean Value Theorem also has a very natural interpretation if $f(x)$ represents the position of an object at time $x$: $f'(x)$ represents the velocity of the object at the instant $x$ and $\displaystyle \frac{f(b) - f(a)}{b - a} = \frac{\mbox{change in position}}{\mbox{change in time}}$ represents the average (mean) velocity of the object during the time interval from time $a$ to time $b$. The Mean Value Theorem says that there is a time $c$ (between $a$ and $b$) when the instantaneous velocity, $f'(c)$, is equal to the average velocity for the entire trip, $\displaystyle \frac{f(b) - f(a)}{b - a}$. If your average velocity during a trip is 30 miles per hour, then at some instant during the trip you were traveling exactly 30 miles per hour.

Practice $\PageIndex{2}$

For $f(x) = 5x^2 - 4x + 3$ on the interval $[1,3]$, calculate $\displaystyle m = \frac{f(b) - f(a)}{b - a}$ and find the value(s) of $c$ so that $f'(c) = m$.

Answer

With $f(x) = 5x^2- 4x + 3$ on $[1, 3]$, $f(1) = 4$ and $f(3) = 36$ so:\[m = \frac{f(b) - f(a)}{b - a} = \frac{36 - 4}{3 - 1} = 16\nonumber\]\)f'(x) = 10x - 4\) so $f'(c) = 10c - 4 = 16 \Rightarrow 10c = 20 \Rightarrow c = 2$. The graph of $f$ showing the location of $c$ appears below:

$fig302_16.png$

Some Consequences of the Mean Value Theorem

If the Mean Value Theorem was just an isolated result about the existence of a particular point $c$, it would not be very important or useful. However, the Mean Value Theorem is the basis of several results about the behavior of functions over entire intervals, and it is these consequences that give it an important place in calculus for both theoretical and applied uses.

The next two corollaries are just the first of many results that follow from the Mean Value Theorem.

We already know, from the Main Differentiation Theorem, that the derivative of a constant function $f(x) = K$ is always $0$, but can a non-constant function have a derivative that is always $0$? The first corollary says no.

Corollary $\PageIndex{1}$

If: $f'(x) = 0$ for all $x$ in an interval $I$

then: $f(x) = K$, a constant, for all $x$ in $I$.

Proof

Assume $f'(x) = 0$ for all $x$ in an interval $I$. Pick any two points $a$ and $b$ (with $a \neq b$) in the interval. Then, by the Mean Value Theorem, there is a number $c$ between $a$ and $b$ so that $\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$. By our assumption, $f'(x) = 0$ for all $x$ in $I$, so we know that $\displaystyle 0 = f'(c) = \frac{f(b) - f(a)}{b - a}$ and thus $f(b) - f(a) = 0 \Rightarrow f(b) = f(a)$. But $a$ and $b$ were two arbitrary points in $I$, so the value of $f(x)$ is the same for any two values of $x$ in $I$, and $f$ is a constant function on the interval $I$.

We already know that if two functions are “parallel” (differ by a constant), then their derivatives are equal, but can two non-parallel functions have the same derivative? The second corollary says no.

Corollary $\PageIndex{2}$

If: $f'(x) = g'(x)$ for all $x$ in an interval $I$

then: $f(x) - g(x) = K$, a constant, for all $x$ in $I$, so the graphs of $f$ and $g$ are “parallel” on the interval $I$

Proof

This corollary involves two functions instead of just one, but we can imitate the proof of the Mean Value Theorem and introduce a new function $h(x) = f(x) - g(x)$. The function $h$ is differentiable and $h'(x) = f'(x) - g'(x) = 0$ for all $x$ in $I$ so, by Corollary 1, $h(x)$ is a constant function and $K = h(x) = f(x) - g(x)$ for all $x$ in the interval. Thus $f(x) = g(x) + K$.

We will use Corollary $\PageIndex{2}$ hundreds of times in Chapters 4 and 5 when we work with “integrals.” Typically you will be given the derivative of a function, $f'(x)$, and be asked to find all functions $f$ that have that derivative. Corollary 2 tells us that if we can find one function $f$ that has the derivative we want, then the only other functions that have the same derivative are of the form $f(x) + K$ where $K$ is a constant: once you find one function with the right derivative, you have essentially found all of them.

Example $\PageIndex{2}$

Find all functions whose derivatives equal $2x$.
Find a function $g(x)$ with $g'(x) = 2x$ and $g(3) = 5$.

Solution

Observe that $f(x) = x^2 \Rightarrow f'(x) = 2x$, so one function with the derivative we want is $f(x) = x^2$. Corollary 2 guarantees that every function $g$ whose derivative is $2x$ has the form $g(x) = f(x) + K = x^2 + K$.
Because $g'(x) = 2x$, we know that $g$ must have the form $g(x) = x^2 + K$, but this gives a whole “family” of functions:
$Graphs of y=x^2-4 (in red, with a black dot at the point (3,5)), y-x^2 (in blue), y=x^2+2 (in pink) and y=x^2+6 (in brown) on a [-2,3]X[-5,10] grid.$
and we want to find one member of that family. We also know that $g(3) = 5$ so we want to find the member of the family that passes through the point $(3,5)$. Replacing $g(x)$ with $5$ and $x$ with $3$ in the formula $g(x) = x^2 + K$, we can solve for the value of $K$: $5 = g(3) = (3)^2 + K \Rightarrow K = - 4$. The function we want is $g(x) = x^2 - 4$.

Practice $\PageIndex{3}$

Restate Corollary $\PageIndex{2}$ as a statement about the positions and velocities of two cars.

Answer: If two cars have the same velocities during an interval of time (so that $f'(t) = g'(t)$ for $t$ in $I$) then the cars are always a constant distance apart during that time interval. (Note: “Same velocity” means same speed and same direction. If two cars are traveling at the same speed but in different directions, then the distance between them changes and is not constant.)

Problems

In the figure below, find the number(s) “$c$” that Rolle’s Theorem promises (guarantees).
$A red curve (labeled f(x)) starts and ends at two blue dots in the first quadrant. The curve starts near x = 1, rises to a peak near x = 2.5, falls steadily to (8,0), where it becomes concave down, then concave up near x = 9, eaching a low point at x = 10 before rising through (12,0), where it becomes concave downm to a peak near x = 11 and falling slightly lower to end at x = 12.$

For Problems 2–4, verify that the hypotheses of Rolle’s Theorem are satisfied for each of the functions on the given intervals, and find the value of the number(s) “$c$” that Rolle’s Theorem promises.

1. $f(x) = x^2$ on $[ -2, 2]$
2. $f(x) = x^2- 5x + 8$ on $[ 0, 5]$
1. $f(x) = \sin(x)$ on $[ 0, \pi]$
2. $f(x) = \sin(x)$ on $[\pi, 5\pi]$
1. $f(x) = x^3- x + 3$ on $[ -1, 1]$
2. $f(x) = x \cdot \cos(x)$ on $[ 0, \frac{\pi}{2}]$
Suppose you toss a ball straight up and catch it when it comes down. If $h(t)$ is the height of the ball $t$ seconds after you toss it, what does Rolle’s Theorem say about the velocity of the ball? Why is it easier to catch a ball that someone on the ground tosses up to you on a balcony, than for you to be on the ground and catch a ball that someone on a balcony tosses down to you?
If $\displaystyle f(x) = \frac{1}{x^2}$, then $f(-1) = 1$ and $f(1) = 1$ but $\displaystyle f'(x) = -\frac{2}{x^3}$ is never equal to $0$. Why doesn't this function violate Rolle’s Theorem?
If $f(x) = \left| x \right|$, then $f(-1) = 1$ and $f(1) = 1$ but $f'(x)$ is never equal to $0$. Why doesn’t this function violate Rolle’s Theorem?
If $f(x) = x^2$, then $f'(x) = 2x$ is never $0$ on the interval $[ 1, 3]$. Why doesn’t this function violate Rolle’s Theorem?
If I take off in an airplane, fly around for a while and land at the same place I took off from, then my starting and stopping heights are the same but the airplane is always moving. Why doesn’t this violate Rolle’s theorem, which says there is an instant when my velocity is $0$?
Prove the following corollary of Rolle’s Theorem: If $P(x)$ is a polynomial, then between any two roots of $P$ there is a root of $P'$.
Use the corollary in Problem 10 to justify the conclusion that the only root of $f(x) = x^3 + 5x - 18$ is $2$. (Suggestion: What could you conclude about $f'$ if $f$ had another root?)
In the figure below, find the location(s) of the “$c$” that the Mean Value Theorem promises:
$A blue curve (labeled y=f(x)) starts and ends at two blue dots. The curve begins concave down at (1,-2), reaching a low point near (2.5,-3), rising through (6,0), where it becomes concave down, reached a peak near (9,4.5), falls through (11,0), where it becomes concave up, reaches a low point near (12.8,1), and rises to (14,2).$

In Problems 13–15, verify that the hypotheses of the Mean Value Theorem are satisfied for each of the functions on the given intervals, and find the number(s) “$c$” that the Mean Value Theorem guarantees.

1. $f(x) = x^2$ on $[ 0, 2]$
2. $f(x) = x^2 - 5x + 8$ on $[ 1, 5]$
1. $f(x) = \sin(x)$ on $[ 0, \frac{\pi}{2}]$
2. $f(x) = x^3$ on $[-1, 3]$
1. $f(x) = 5 - \sqrt{x}$ on $[1, 9]$
2. $f(x) = 2x + 1$ on $[ 1, 7]$
For the quadratic functions in parts (a) and (b) of Problem 13, the number $c$ turned out to be the midpoint of the interval: $c = \frac{a + b}{2}$.
1. For $f(x) = 3x^2 + x - 7$ on $[ 1, 3]$, show that $\displaystyle f'(2) = \frac{f(3) - f(1)}{3 - 1}$.
2. For $f(x) = x^2 - 5x + 3$ on $[ 2, 5]$, show that $\displaystyle f'\left( \frac72 \right) = \frac{f(5) - f(2)}{5 - 2}$.
3. For $f(x) = Ax^2 + Bx + C$ on $[ a, b]$, show that $\displaystyle f'\left( \frac{a+b}{2}\right) = \frac{f(b) - f(a)}{b - a}$.
If $f(x) = \left| x \right|$, then $f(-1) = 1$ and $f(3) = 3$ but $f'(x)$ is never equal to $\displaystyle \frac{f(3) - f(-1)}{3 - (-1)} = \frac12$. Why doesn’t this violate the Mean Value Theorem?

In Problems 18–19, you are a traffic-court judge. In each case, a driver has challenged a speeding ticket and you need to decide if the ticket is appropriate.

A tolltaker says, “Your Honor, based on the elapsed time from when the car entered the toll road until the car stopped at my booth, I know the average speed of the car was 83 miles per hour. I did not actually see the car speeding, but I know it was and I gave the driver a ticket.”
The driver in the next case heard the tolltaker and says, “Your Honor, my average velocity on that portion of the toll road was only 17 miles per hour, so I could not have been speeding.”
Find three different functions ($f$, $g$ and $h$) so that $f'(x) = g'(x) = h'(x) = \cos(x)$.
Find a function $f$ so that $f'(x) = 3x^2 + 2x + 5$ and $f(1) = 10$.
Find $g(x)$ so that $g'(x) = x^2 + 3$ and $g(0) = 2$.
Find values for $A$ and $B$ so that the graph of the parabola $f(x) = Ax^2 + B$ is:
1. tangent to $y = 4x +5$ at the point $(1,9)$.
2. tangent to $y = 7 - 2x$ at the point $(2,3)$.
3. tangent to $y = x^2 + 3x - 2$ at the point $(0,2)$.
Sketch the graphs of several members of the “family” of functions whose derivatives always equal $3$. Give a formula that defines every function in this family.
Sketch the graphs of several members of the “family” of functions whose derivatives always equal $3x^2$. Give a formula that defines every function in this family.
At $t$ seconds after takeoff, the upward velocity of a helicopter was $v(t) = 3t^2+2t$ feet/second. Two seconds after takeoff, the helicopter was 80 feet above sea level. Find a formula for the height of the helicopter at every time $t$.
Assume that a rocket is fired from the ground and has the upward velocity shown in the figure below:
$A [0,5.5]X[0,350] grid is a labeled 'time (seconds)' on the horizontal axis and 'velocity (ft/sec)' on the vertical axis. A red line segment extends from (0,300) to (5,300).$
Estimate the height of the rocket when $t = 1$, $2$ and $5$ seconds.
The figure below shows the upward velocity of a rocket:
$A [0,5.5]X[0,420] grid is a labeled 'time (seconds)' on the horizontal axis and 'velocity (m/sec)' on the vertical axis. A blue, piecewise-linear graph starts passes through (0,200), (1,200), (2,300), (4,300) and (5,400).$
Use the information in the graph to estimate the height of the rocket when $t = 1$, $2$ and $5$ seconds.
Determine a formula for $f(x)$ if you know: $f''(x) = 6$, $f'(0) = 4$ and $f(0) = -5$.
Determine a formula for $g(x)$ if you know: $g''(x) = 12x$, $g'(1) = 9$ and $g(2) = 30$.
Define $A(x)$ to be the area bounded by the $t$-axis, the line $y = 3$ and a vertical line at $t = x$:
1. Find a formula for $A(x)$.
2. Determine $A'(x)$.
Define $A(x)$ to be the area bounded by the $t$-axis, the line $y = 2t$ and a vertical line at $t = x$:
1. Find a formula for $A(x)$.
2. Determine $A'(x)$.
Define $A(x)$ to be the area bounded by the $t$-axis, the line $y = 2t+1$ and a vertical line at $t = x$:
1. Find a formula for $A(x)$.
2. Determine $A'(x)$.

In Problems 34–36, given a list of numbers $a_1$, $a_2$, $a_3$, $a_4$, … , the consecutive differences between numbers in the list are: $a_2 - a_1$, $a_3 - a_2$, $a_4 - a_3$, …

If $a_1 = 5$ and the consecutive difference is always $0$, what can you conclude about the numbers in the list?
If $a_1 = 5$ and the consecutive difference is always $3$, find a formula for $a_n$.
Suppose the “$a$” list starts with $3$, $4$, $7$, $8$, $6$, $10$, $13$, … , and there is a “$b$” list that has the same consecutive differences as the “$a$” list.
1. If $b_1 = 5$, find the next six numbers in the “$b$” list. How is $b_n$ related to $a_n$?
2. If $b_1 = 2$, find the next six numbers in the “$b$” list. How is $b_n$ related to $a_n$?
3. If $b_1 = B$, find the next six numbers in the “$b$” list. How is $b_n$ related to $a_n$?

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Rolle’s Theorem

Proof

Example \(\PageIndex{1}\)

Solution

Practice \(\PageIndex{1}\)

Mean Value Theorem

Proof

Practice \(\PageIndex{2}\)

Corollary \(\PageIndex{1}\)

Proof

Corollary \(\PageIndex{2}\)

Proof

Example \(\PageIndex{2}\)

Solution

Practice \(\PageIndex{3}\)