Statement of Mean Value Theorem MVT : Let a function [katex]f[/katex] be
- contineous on [katex]\left[a\;,\;b\right][/katex].
- differentiable on [katex]\rbrack a,b\lbrack [/katex].
then there exist a point [katex]c\in\rbrack a,b\lbrack[/katex] such that
[katex]\frac{f(b)-f(a)}{b-a}=\;f'(c)[/katex]
MVT Proof: Define a new function [katex]F[/katex] by
[katex]F(x)= Ax + f(x)[/katex]
where A is a constant to determined such that [katex]F(a)=F(b)[/katex]
The functiom [katex]Ax[/katex] is contineous on [katex]\left[a\;,\;b\right] [/katex] and differentiable on [katex]\rbrack a,b\lbrack [/katex]. Hence [katex]F(x)[/katex] is contineous and differentiable respecctively on [katex]\left[a\;,\;b\right] [/katex] and [katex]\rbrack a,b\lbrack [/katex]. The function [katex]F[/katex] satisfies all the conditions of Rolle’s theorem therefore, there exist a point [katex]c\in\rbrack a,b\lbrack[/katex] such that
[katex]F'(c)=0\;\;\;\; i.e,\;\;\; A + f'(c)=0 [/katex]
or [katex]f'(c)=-A[/katex]
From [katex]F(a)=F(b)[/katex], we have
[katex]Aa+f(a)=Ab +f(b)[/katex]
or [katex]-A(b-a) =f(b)- f(a) [/katex]
or [katex]-A=\frac{f(b)-f(a)}{b-c}[/katex]
substituting this value of [katex]A[/katex] into [katex](1)[/katex], we get
[katex]\frac{f(b)-f(a)}{b-c}=f'(c)[/katex] as required.
Geometrical Interpretation of Mean Value Theorem, MVT
The mean value theorem MVT has asimple geometrical interpretation
Let the graph of f on [katex]\left[a\;,\;b\right][/katex] represented by the curve APB. Suppose the chord AB makes an angle of measure [katex]\beta[/katex] with the x axis
Then
[katex]\tan\;\beta\;=\frac{BM}{AM}=\frac{f(b)-f(a)}{b-a}……………(1)[/katex]
by the mean value theorem, we have
[katex]\frac{f(b)-f(a)}{b-a}=f'(c)……………(2)[/katex]
where [katex]c\in\rbrack a,b\lbrack[/katex]
From [katex](1)[/katex] and [katex](2)[/katex] we get [katex]\tan\;\beta\;=f'(c)[/katex]
Thus at the point [katex]P (c, f(c) )[/katex], The tangent line ton the graph of [katex]y=f(x)[/katex] is parallel to the chord of [katex]AB[/katex].
For physical interpretation of MVT , Suppose [katex]s= f(t)[/katex] denotes the distance S that a particle has travelled at time [katex]t[/katex]. If [katex]f[/katex] contineous on [katex]\left[a\;,\;b\right][/katex]. and differentiable on [katex]\rbrack a,b\lbrack [/katex] then by mean value theorem (mvt) there exists a point [katex]t_0\in\rbrack a,b\lbrack[/katex] such that
[katex]\frac{f(b)-f(a)}{b-c}=f'(t_0)[/katex]
[katex] i.e,[/katex] the velocity [katex]f'(t_0)[/katex] equal to the average velocity [katex]\frac{f(b)-f(a)}{b-a}[/katex] from [katex]a[/katex] to [katex]b[/katex].
For Example, If a car travels 400km in 5 hours then at some time the car had the speed of 80km per hour.
Also called LaGrange’s Mean Value Theorem after the name of French Mathematician J.L Lagrange(1736-1813)