Verify Rolle’s theorem for the function f (x) = 
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) = 
exists for all x Î (–2, 2)
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) = 
Þ f ¢ (x) =
Þ f ¢ (x) = \ f ¢ (x) = 0 Þ = 0 Þ x = 0
= 0 Þ x = 0Since c = 0 Î (– 2, 2) such that f ¢ (c) = 0
Hence, Rolle’s theorem is verified.
Answered by
| 4th Aug, 2014,
05:32: PM
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