Verify Rolle's theorem for the function f (x) = log space open curly brackets fraction numerator straight x squared plus ab over denominator straight a left parenthesis straight a plus straight b right parenthesis end fraction close curly brackets on [a, b], where 0 < a < b.

Asked by Topperlearning User | 4th Aug, 2014, 03:48: PM

Expert Answer:

We have,

f (x) = log space open curly brackets fraction numerator straight x squared plus ab over denominator straight x left parenthesis straight a plus straight b right parenthesis end fraction close curly brackets = 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 open curly brackets fraction numerator a squared plus a b over denominator a left parenthesis a plus b right parenthesis end fraction close curly brackets = log 1 = 0, and f (b) = log open curly brackets fraction numerator b squared plus a b over denominator b left parenthesis a plus b right parenthesis end fraction close curly brackets = 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) = fraction numerator 2 straight x over denominator straight x squared plus ab end fraction minus 1 over straight x equals fraction numerator straight x squared minus ab over denominator straight x left parenthesis straight x squared plus ab right parenthesis end fraction
\ f '(x) = 0 & fraction numerator straight x squared minus ab over denominator straight x left parenthesis straight x squared plus ab right parenthesis end fraction= 0 =>x2 = ab => xsquare root of ab
Since a<square root of abc = square root of ab ε (a, b) is such that f '(c) = 0
Hence, Rolle's theorem is verified.

Answered by  | 4th Aug, 2014, 05:48: PM