CBSE Class 12-science Answered
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, 15:57: PM
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) =
![limit as straight x rightwards arrow 0 to the power of minus of fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction](https://images.topperlearning.com/topper/tinymce/cache/ebbd093f773b917adc3c143df2619b38.png)
Þ (LHD at x = 0) =
[Q f (x) = – x for x < 0 and f (0) = 0]
![limit as straight x rightwards arrow 0 of fraction numerator minus straight x minus 0 over denominator straight x end fraction equals minus 1](https://images.topperlearning.com/topper/tinymce/cache/5a97996ac431e99462ec1b1eb433d944.png)
and, (RHD at x = 0) =
[Q f (x) = x for x ³ 0]
![limit as straight x rightwards arrow 0 to the power of plus of fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals limit as straight x rightwards arrow 0 to the power of minus of fraction numerator straight x minus 0 over denominator straight x end fraction equals 1](https://images.topperlearning.com/topper/tinymce/cache/5381c9920742679a2b7c4cc045591be9.png)
\ (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, 17:57: PM
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