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Differential Calculus

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17 Mar 2008 20:21:02 IST

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A beaut

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$\text{Let f(x) be continuous on [a,b] and differentiable on (a,b)} \\ \\ \text{Prove that there exists} \ c \in (a,b) \ \text {such that} \\ \\ f$

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sandeep ramesh

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17 Mar 2008 21:02:20 IST

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

Nadeem

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17 Mar 2008 21:12:34 IST

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@ sandeep

put c=1.5

sandeep ramesh

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17 Mar 2008 21:17:44 IST

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sorry i wrote that like a fool

Somebody kick me

Anand Hegde

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17 Mar 2008 21:22:04 IST

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take that!

Nadeem

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17 Mar 2008 21:22:54 IST

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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)

sandeep ramesh

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17 Mar 2008 21:29:28 IST

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

Anand Hegde

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17 Mar 2008 21:33:27 IST

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Hari Shankar

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17 Mar 2008 21:37:47 IST

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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.

Abhijith

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17 Mar 2008 22:17:23 IST

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Nice question sir!

Let

Then

therefore there exists

such that

But

hence

which gives you:

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