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Discuss the applicability of Rolle’s theorem for the following function on the indicated interval: f(x) = |x| on [–1, 1]
Asked by Topperlearning User | 04 Aug, 2014, 03:57: PM Expert Answer

We have,    f (x) = |x| = Since a polynomial function is everywhere continuous and differentiable. Therefore, f (x) is continuous and differentiable for all x < 0 and for all x > 0 except possibly at x = 0.
So, consider the point x = 0.
We have, Thus, f (x) is continuous at x = 0
Hence, f (x) is continuous on [–1, 1]
Now,
(LHD at x = 0) = Þ (LHD at x = 0) = [Q f (x) = – x for x < 0 and f (0) = 0]
and, (RHD at x = 0) = [Q f (x) = x for x ³ 0]

\ (LHD at x = 0) ¹ RHD at x = 0
This shows that f (x) is not differentiable at x = 0 Î (– 1, 1). Thus, the condition of derivability at each point of (– 1, 1) is not satisfied.
Hence, Rolle’s theorem is not applicable to f (x) = |x| on [–1, 1]
Answered by | 04 Aug, 2014, 05:57: PM
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