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Consider two police officers. One officer is parked with a radar gun next to mile marker 8, while the other is parked with a radar gun at mile marker 17. A car passes the first officer at 1:00 pm. The radar gun says the vehicle travels at 65 mph. Nine miles down, the car passes the second officer at 1:06 pm at 67 mph. The speed limit is 70 mph. Should the officers issue the vehicle a speeding ticket?
Why would the officers issue a speeding ticket if the car was clocked traveling below the speed limit at both mile markers? Let's focus on the more important question: How did the car travel 9 miles in 6 minutes? This would mean that the car travels with an average speed of 90 mph which is certainly over the speed limit. The Mean Value Theorem is a Calculus theorem that ensures the car could not possibly have an average speed of 90 mph without traveling at exactly 90 mph at least once between the two police officers.
The Mean Value Theorem highlights a link between the tangent and secant lines. Although the result may seem somewhat obvious, the theorem is used to prove many other theorems in Calculus.
The Mean Value Theorem states that if a function f is:
continuous on the closed interval [a, b]
differentiable on the open interval (a, b)
then there is a number c such that a < c < b and
In words, the theorem states that for at least one point on the curve between endpoints a and b, the slope of the tangent line (instantaneous rate of change) will be equal to the slope of the secant line (average rate of change) through the points (a, f(a)) and (b, f(b)).
The slope of the tangent line between two points must equal the slope of the secant line between those two points at least once according to the Mean Value Theorem - StudySmarter Original
In a real-world application, the Mean Value Theorem says that if you drive 40 miles in one hour, then at some point within that hour, your speed will be exactly 40 miles per hour.
Let A represent the point (a, f(a)) and B represent the point (b, f(b)).
Notice how the right-hand side of the Mean Value Theorem is the slope of the secant line through points A and B. Using point A to form the equation for the secant line, we get
Let F(x) be the vertical distance between some point (x, f(x)) on the graph of f and the corresponding point on the secant line through A and B. Then, F is positive when the graph of f is above the secant line and negative otherwise. In mathematical terms,
We want to show that F satisfies the conditions of Rolle's Theorem. For more information on this theorem, please view our article on Rolle's Theorem.
With the conditions of Rolle's Theorem satisfied, we know that some c in the interval (a, b) must exist such that .
When we differentiate F(x), we find
Using our conclusion from satisfying Rolle's Theorem,
So, there exists some c in the interval (a, b) such that
Let's look at a real-world application of the Mean Value Theorem.
A ball is dropped from a height of 100 ft. Its position in seconds after it's dropped is modeled by the function .
Before using the Mean value Theorem, we must ensure our function meets the theorem's requirements.
Since is a polynomial, we know it is continuous and differentiable on the entire interval!
The ball is dropped from a height of 100 ft. Let the ground be at 0 ft. So, to find when the ball hits the ground, we can set and solve for
!
As our time in seconds cannot be negative, the ball must hit the ground at , or 2.5, seconds.
We will use the time the ball is released, , and the time the ball hits the ground
(which is the total time that the ball is in free fall) to find the average velocity of the ball over the course of its fall. Averaging over both values...
Therefore, the average velocity of the ball is (down) during the time it is in the air. The sign of the velocity is negative here because the ball travels in the negative (in this case, down) direction.
The Mean Value Theorem states that there is at least one point on seconds where the ball has an instantaneous velocity of
(down).
We'll start by taking the derivative of the position function s(t).
So, the ball reaches a velocity of -40 ft per second at time , or 1.25, seconds.
Assume that a function f(x) is continuous and differentiable on the interval [5, 15]. Given f(5) = 4 and f'(x)10. Find the largest possible value that f(15) can take on.
The problem specifies that f(x) is in fact continuous and differentiable. Thus, we can apply the Mean Value Theorem.
We are looking for the value of f(15), or f(b). So, let's rearrange the equation to solve for f(b).
If f'(x) has a maximum value of 10, then f'(c) also has a maximum value of 10. So, we can substitute 10 in for f'(c) into the Mean Value Theorem.
So, the maximum value that f(15) can be is 104.
The Mean Value Theorem can also be applied specifically to integrals. Details on this can be found in our article The Mean Value Theorem for Integrals.
The Mean Value Theorem illustrates the like between the tangent line and the secant line. For at least one point on the curve between endpoints a and b, the slope of the tangent line will be equal to the slope of the secant line through the point (a, f(a)) and (b, f(b)).
The Mean Value Theorem is used to prove a multitude of other theorems in Calculus.
The Mean Value Theorem can be verified by making sure the the derivative at some point c is equal to the average value of the function on the interval (a, b)
Unknown values can be found by plugging in known values to the Mean Value Theorem formula and solving for the unknown value.
The Mean Value Theorem is used in applications of position and velocity. It can also be used to prove a multitude of other Calculus theorems.
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