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, 03:57: PM

Expert Answer

We have, f (x) = |x| = $open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell minus straight x comma end cell cell when space minus 1 less or equal than straight x less or equal than 0 end cell row cell straight x comma end cell cell when space 0 less or equal than straight x less or equal than 1 end cell end table close$

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,

$limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis equals limit as straight x rightwards arrow 0 of minus straight x equals 0 space and space limit as straight x rightwards arrow 0 to the power of plus of straight f left parenthesis straight x right parenthesis equals limit as straight x rightwards arrow 0 to the power of minus of straight x equals 0 therefore limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis equals limit as straight x rightwards arrow 0 plus of straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis$

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

Þ (LHD at x = 0) = _{$limit as straight x rightwards arrow 0 of fraction numerator minus straight x minus 0 over denominator straight x end fraction equals minus 1$} [Q f (x) = – x for x < 0 and f (0) = 0]

and, (RHD at x = 0) = _{$limit as straight x rightwards arrow 0 to the power of plus of fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals limit as straight x rightwards arrow 0 to the power of minus of fraction numerator straight x minus 0 over denominator straight x end fraction equals 1$} [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