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Verify Rolle's theorem for the function f (x) = on [a, b], where 0 < a < b.
Asked by Topperlearning User | 04 Aug, 2014, 03:48: PM Expert Answer

We have,

f (x) = = log (x2 + ab) - log x - log (a + b)
Since logarithimic function is differentiable and so continuous on its domain. Therefore, f (x) is continuous on [a, b] and differentiable on (a, b).
Also, f (a) = = log 1 = 0, and f (b) = = log 1 = 0
\ f (a) = f (b)
Thus, all the three conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists ε (a, b) such that f '(c) = 0
We have,
f (x) = log (x2 + ab) - log x - log (a + b)
f '(x) = \ f '(x) = 0 & = 0 =>x2 = ab => x Since a< <b Therefore, c = ε (a, b) is such that f '(c) = 0
Hence, Rolle's theorem is verified.
Answered by | 04 Aug, 2014, 05:48: PM

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