Verify Rolle’s theorem for the function f (x) = square root of 4 minus straight x squared end root on [– 2, 2].

Asked by Topperlearning User | 4th Aug, 2014, 03:32: PM

Expert Answer:

Clearly, f (x) is defined for all x Î [– 2, 2] and has a unique value for each x Î [– 2, 2]. So, at each point of [–2, 2], the limit of f (x) is equal to the value of the function. Therefore, f (x) is continuous on [–2, 2].

Also, f ¢ (x) = fraction numerator minus straight x over denominator square root of 4 minus straight x squared end root end fraction exists for all x Î (–2, 2)
So, f (x) is differentiable on (–2, 2)
Also, f (–2) = f (2) = 0
Thus, all the three conditions of Rolle’s theorem are satisfied.
Now we have to show that there exists c Î (–2, 2) such that f ¢ (c) = 0
We have,
f (x) = square root of 4 minus straight x squared end root Þ f ¢ (x) = fraction numerator minus straight x over denominator square root of 4 minus straight x squared end root end fraction
\ f ¢ (x) = 0 Þ fraction numerator minus straight x over denominator square root of 4 minus straight x squared end root end fraction = 0 Þ x = 0
Since c = 0 Î (– 2, 2) such that f ¢ (c) = 0
Hence, Rolle’s theorem is verified.

Answered by  | 4th Aug, 2014, 05:32: PM