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Gramin ACS College Vasantnagar, Kotgyal � MATHEMATIS

B.Sc. Second Year, Paper-I (Differencial calculus)�Sem-I�PPT Presented by.....�Prof. Dr. P. R. Shinde�Department of Mathematics�Gramin ACS college Vasantnagar, kotgyal�Tq: Mukhed Dist: Nanded ( M.S) INDIA�

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Mean value theorem

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� Introduction:

By now the student must have learnt to distinguish between theorems applicable to a class of function and those concerning some particular function like f(x)=sinhx, f(x)=log x etc.

The theorems applicable to a class of function are known as general.

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ROLL’S THEOREM : If a function f is:

(I). Continuous in a closed internal [ a,b ]
(II). Derivable in the open internal ] a,b [ and
f(a) =f(b), then there exists at least one value ‘C’ € ]a,b[ (II). Derivable in the open internal ] a,b [ and
PROOF : The continuity of the function in the closed internal [ a,b ] it has a greatest value M and a least value m in the internal, so that there are two numbers c and d such that f(c)=M, f(d)=m.
Now either M=m................(1)
Or 𝑀≠𝑚 .............(2)
When the greatest value coincides with the least value as in case.

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(I) The function reduces to a constant so that the derivative is equal to 0 for every value of x and the theorem is true in this case

(II) When M and m are unequal, as in case(II) at least one of the must be different from the equal values. f(a), f(b). Let M=f(c) be different from a and b lies within the internal [ a,b ] and as such belongs to the open internal ] a,b [ is in particular Derivable at c , so that lim⁡〖(𝑓(𝑐+ℎ)−𝑓(𝑐))/ℎ〗 when h->0.

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Also f(c) is the greatest value of the function, we have 𝑓(𝑐+ℎ)≤𝑓(𝑐)
Whatever positive or negative value h has.
Thus (𝑓(𝑐+ℎ)−𝑓(𝑐))/ℎ ≤0 𝑓𝑜𝑟 ℎ>0………(3)
𝑓(𝑐+ℎ)−𝑓(𝑐)/ℎ≥0 𝑓𝑜𝑟 ℎ<0……..(4)
Let h->0 through positive value from equation (3) we get f’(c) ≤ 0..........(5)
Let h->0 through negative equation (4) we get f’(c) ≥ 0............(6)
The relation equation (5) & (6) will be the TRUE if and only if f’(c)=0.

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Examples

(1). Verify ROLL’S theorem for f(x)=𝑥^2 in [ -1, 1]

ANS : LET f(a) = 𝑥^2 , x€[ -1, 1]

So that f(1)= 1=f(-1)

Also the function is Derivable in [-1,1]

The condition of the theorem being satisfied, the derivable f’(x) must vanish for at least one value of x € ]-1, 1 [.

Directly, we see that the derivative vanishes for x=0 which belongs to ] -1,1 [

Hence proved.