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Mean Value Theorem
Mean Value Theorem is an important theorem in calculus. The first form of the mean value theorem was proposed in the 14th century by Parmeshwara, a mathematician from Kerela, India. Further, a simpler version of this was proposed by Rolle in the 17th century: Rolle’s Theorem, which was proved only for polynomials and was not a part of the calculus. Finally, the present version of the Mean Value Theorem was proposed by Augustin Louis Cauchy in the year 1823.
The mean value theorem states that for a curve passing through two given points there is one point on the curve where the tangent is parallel to the secant passing through the two given points. Rolle’s theorem has been derived from this mean value theorem.
What is Mean Value Theorem?
The mean value theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing through the two given points. The mean value theorem is defined herein calculus for a function f(x): [a, b] → R, such that it is continuous and differentiable across an interval.
- The function f(x) is continuous across the interval [a, b].
- The function f(x) is differentiable across the interval (a, b). There exists a point c in (a, b).
Here we have proved that the tangent at c is parallel to the secant passing through the points (a, f(a)), (b, f(b)). This mean value theorem is used to prove a statement across a closed interval. Further, Rolle’s theorem is derived from this mean value theorem.
Proof of Mean Value Theorem
Statement: The mean value theorem states that if a function f is continuous over the closed interval [a,b], and differentiable over the open interval (a,b), then there exists at least one point c in the interval (a,b) such that f(c) is the average rate of change of the function over [a,b] and it is parallel to the secant line over [a,b].
Proof: Let g(x) be the secant line to f(x) passing through (a, f(a)) and (b, f(b)). We know that the equation of the secant line is y-y1 = m (x- x1).
Thus the mean value theorem is proved.
Note: The result may not hold if the function is not differentiable, even at a single point.
Graphical Representation of Mean Value Theorem
The graphical representation of the function f(x) helps in understanding the mean value theorem. Here we consider two distinct points (a, f(a)), (b, f(b)). The line connecting these points is the secant of the curve, which is parallel to the tangent cutting the curve at (c, f(c)). The slope of the secant of the curve joining these points is equal to the slope of the tangent at the point (c, f(c)). We know that the derivative of the tangent is the slope at that point.
Slope of the Tangent = Slope of the Secant
Here we observe that the point (c, f(c)), lies between the two points (a, f(a)), (b, f(b)).
Difference Between Mean Value Theorem and Rolle’s Theorem
The difference between the mean value theorem and Rolle’s theorem helps in a better understanding of these theorems. Both the theorems define the function f(x) such that it is continuous across the interval [a, b], and it is differentiable across the interval (a, b). In the mean value theorem, the two referred points (a, f(a)), (b, f(b)) are distinct and f(a) ≠ f(b). In Rolle’s theorem, the points are defined such that f(a) = f(b).
The value of c in the mean value theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the secant joining the two points. The value of c in Rolle’s theorem is defined such that the slope of the tangent at the point (c, f(c)) is equal to the slope of the
Examples of Mean Value Theorem
Example 1: Verify if the function f(x) = x2 + 1 satisfies mean value theorem in the interval [1, 4].
Solution:
The given function is f(x) = x2 + 1. To verify the mean value theorem, the function f(x) = x2 + 1 must be continuous in [1, 4] and differentiable in (1, 4).
The derivative f'(x) = 2x is defined in the interval (1, 4)
f(1) = 12 + 1 = 1 + 1 = 2
f(4) = 42 + 1 = 16 + 1 = 17
f'(x) = 5
2x = 5
x = 2.5
Answer: Since 2.5 lies in the interval (1, 4), the function satisfies the mean value theorem.
Example 2: Find the value of c if the function f(x) = x2 – 4x + 3 satisfies mean value theorem in the interval (1, 4).
Solution:
The given function f(x) = x2 – 4x + 3 satisfies mean values theorem. Hence it is continuous in [1, 4] and is differentiable in (1, 4).
f'(x) = 2x – 4
f(1) = 1 – 4 + 3 = 0
f(4) = 42 – 4(4) + 3 = 16 – 16 + 3 = 3
f'(c) = 1
2c – 4 = 1
2c = 5
c = 5/2 = 2.5
c = 2.5 belongs to the interval (1, 4)
Answer: c = 2.5
Example 3: For the function f(x) = x2 + 2x, find all the values of c that satisfy the mean value theorem, over the interval [-4,4].
Solution:
f(x) = x2 + 2x is a polynomial and hence it is continuous and differentiable over the given interval [4,-4]
f'(x) = 2x+ 2
f(4) =42 + 2(4) = 24
f(-4) = (-4)2 + 2(-4)= 8
Let us find C in (-4,4) such that f'(c) = 2
f'(x) = 2x+ 2
f'(2) = 2(2)+ 2 = 6
f'(x) = 2c+ 2 = 2
⇒ c = 0.
Answer: For the function f(x) = x2 + 2x, the value of C = 0 that satisfy the mean value theorem, over the interval [-4,4].
FAQs on Mean Value Theorem
What is the Mean Value Theorem?
The mean value theorem states that if a function f is continuous over the closed interval [a,b], and differentiable over the open interval (a,b), then there exists a point c in the interval (a,b) such that f(c) is the average rate of change of the function over [a,b] and it is parallel to the secant line over [a,b].
What is the Mean Value Theorem Equation?
The mean value theorem is defined for a function f(x): [a, b]→ R, such that it is continuous in the interval [a, b], and differentiable in the interval (a,b). For a point c in (a, b), the equation for the mean value theorem is as follows.
What Does Mean Value Theorem Mean?
The mean value theorem states that for a curve passing through two given points there is one point on the curve where the tangent is parallel to the secant passing through the two given points. Rolle’s theorem has been derived from this mean value theorem.
What is the Hypothesis of Mean Value Theorem?
The hypothesis for the mean value theorem is that, for a continuous function f(x), it is continuous in the interval [a, b], and it is differentiable in the interval (a, b).
How to Find the Values that Satisfy Mean Value Theorem?
The values satisfying the mean value theorem are calculated by finding the differential of the given function f(x). The given function is defined in the interval (a, b), and the value satisfying the mean value theorem is the point c, which belongs to the interval (a, b). And we can find its value from
How to Find C for Mean Value Theorem in Integrals?
As per the mean value theorem for the function f(x) defined in the interval (a, b), the value of C belongs to (a, b), and is calculated using the slope of the secant connecting the points (a, f(a)), (b, f(b)). The value of c is calculated from the derivative formula of
Mean Value Theorem Formula
The mean value theorem formula tells us about a point c that must exist in a function if it follows the following conditions:
Let f(x) be a function defined on [a, b] such that
(i) It is continuous on [a, b].
(ii) It is differentiable on (a, b).
Then there exists a real number c∈(a,b), and tangent to the curve of the function at point c will be parallel to the secant line between the points (a, f(a)), and (b, f(b)).
Let us learn more about the mean value theorem formula using solved examples in the section given below.
What is Mean Value Theorem Formula?
If a function follows all the above-mentioned conditions, then there will be a point c between (a, b) where the tangent to the curve of the function at point c will be parallel to the secant line between the points (a, f(a)), and (b, f(b)).
Mean value theorem formula is:
Solved Examples Using Mean Value Theorem Formula
- Example 1: Find the value of c, for which the following function 4x3−8x2+7x−2, satisfies the mean value theorem between the interval [2, 5]. Solution: To find: c
f(2) = 12
f(5) = 333
f′(x)=12×2−16x+7
Using the mean value theorem formula,
12c2−16c+7=107c=−2.29, and 3.629
Since only 3.629 falls under the given interval.
Answer: Hence, the value of c is 3.629.
Example 2:
Determine all the value of c, for the function f(x) = x2−5x+7
for the interval [-1, 3] which satisfies the mean value theorem.
Solution:
To find: c
f(-1) = 13
f(3) = 1
f′(x)=2x−5
Using the mean value theorem formula,
Solved Example
Question: Evaluate f(x) = x2 + 2 in the interval [1, 2] using mean value theorem.
Solution:
Given function is:
f(x) = x2 + 2
Interval is [1, 2].
i.e. a = 1, b = 2
Mean value theorem is given by,
f(b) = f(2) = 22 + 2 = 6
f(a) = f(1) = 12 + 2 = 3
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