Discuss the applicability of Rolle’s theorem on the function f (x) = open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell straight x plus 1 comma end cell cell when space 0 less or equal than straight x less or equal than 1 end cell row cell 3 minus straight x comma end cell cell when space 1 less or equal than straight x less or equal than 2 end cell end table close

Asked by Topperlearning User | 4th Aug, 2014, 04:11: PM

Expert Answer:

Since a polynomial function is everywhere continuous and differentiable. Therefore, f (x) is continuous and differentiable at all points except possibly at x = 1.

Now, we consider the differentiability of f (x) at x = 1.
We have,
(LHD at x = 1) = limit as straight x rightwards arrow 1 to the power of minus of fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator straight x minus 1 end fraction equals fraction numerator left parenthesis straight x squared plus 1 right parenthesis minus left parenthesis 1 plus 1 right parenthesis over denominator straight x minus 1 end fraction equals fraction numerator left parenthesis straight x squared minus 1 right parenthesis over denominator left parenthesis straight x minus 1 right parenthesis end fraction equals left parenthesis straight x plus 1 right parenthesis equals 2
[f (x) = x2 + 1 for 0 ≤ x ≤ 1]
 
(RHD at x = 1) = limit as x rightwards arrow 1 to the power of plus of fraction numerator f left parenthesis x right parenthesis minus f left parenthesis 1 right parenthesis over denominator x minus 1 end fraction equals limit as x rightwards arrow 1 to the power of plus of fraction numerator left parenthesis 3 minus x right parenthesis minus left parenthesis 1 plus 1 right parenthesis over denominator x minus 1 end fraction equals limit as x rightwards arrow 1 to the power of plus of fraction numerator left parenthesis minus x plus 1 right parenthesis over denominator x minus 1 end fraction equals minus 1 
\ (LHD at x = 1) ¹ (RHD at x = 1)
So, f (x) is not differentiable at x = 1.
Thus, the condition of a differentiability at each point of the given interval is not satisfied.
Hence, Rolle’s theorem is not applicable to the given function on the interval [0, 2].

Answered by  | 4th Aug, 2014, 06:11: PM