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 abovementioned 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:
\[f'(c) = \dfrac{f(b)  f(a)}{ba}\]
Solved Examples Using Mean Value Theorem Formula

Example 1:
Find the value of c, for which the following function 4x^{3}−8x^{2}+7x−2, satisfies the mean value theorem between the interval [2, 5].
Solution:
To find: c
f(2) = 12
f(5) = 333
\(f'(x) = 12x^216x+7\)
Using the mean value theorem formula,
\(f'(c) = \dfrac{f(b)  f(a)}{ba}\)
\(\begin{align*} 12c^216c+7 &= \dfrac{33312}{52} \\ 12c^216c+7 &= 107 \\ c &= 2.29, \text{ and } 3.629 \end{align*}\)
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) = \(x^25x+7\) for the interval [1, 3] which satisfies the mean value theorem.
Solution:
To find: c
f(1) = 13
f(3) = 1
\(f'(x) = 2x5\)
Using the mean value theorem formula,
\(f'(c) = \dfrac{f(b)  f(a)}{ba}\)
\(\begin{align} 2c5 &= \dfrac{113}{3(1)} \\ 2c5 &= 3 \\ c &= 1 \end{align}\)
Answer: Hence, the value of c is 1.
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