IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sandeepramesh

umm nice. But how does this work for a = 1, b = 2 and f(x) = 1 for all x?

nadeemoidu

@ sandeep

put c=1.5

sandeepramesh

sorry i wrote that like a fool

Somebody kick me

anandghegde

take that!

nadeemoidu

Consider the function g(x)= rac{1}{a-x} + rac{1}{b-x}

$displaystylelim_{x o a^+}g(x) = - infty$

$displaystylelim_{x o b^-}g(x) = + infty$

Also , g(x) is continuous in (a,b) .

So g(x) takes all the values from ( - infty , infty )

.

So for some x= c in (a,b)

, the curves , f'(x) and g(x) intersect . So f'(c) = g(c)

sandeepramesh

yes perfect man!!! i believe that is absolutely correct

anandghegde

hsbhatt

nadeem, marvellous job yet again. That's nice use of intuitive thinking. With nothing said about f(x), the starting point is to note that we can get all real numbers as g maps (a,b) to R.

Another way is to use the function g(x) = (x-a) (x-b) e^f(x) and then apply Rolle's Theorem as g(a) = g(b) = 0.

konichiwa2x

Nice question sir!

Let

Then

therefore there exists

such that

But

hence

,

which gives you:

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