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  • Discuss the applicability of Rolle’s theorem on the function given by f(x) = x^2+1

Discuss the applicability of Rolle’s theorem on the function given by
f(x)=\begin{array}{ll} x^{2}+1, & \text { if } 0 \leq x \leq 1 \\ 3-x, & \text { if } 1 \leq x \leq 2 \end{array}

Answers (1)

Given: f(x)=\begin{array}{ll} x^{2}+1, & \text { if } 0 \leq x \leq 1 \\ 3-x, & \text { if } 1 \leq x \leq 2 \end{array}

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

At x = 1
\\\lim _{\mathrm{LHL}}\left(\mathrm{x}^{2}+1\right)=1+1=2$ \\$\lim _{\mathrm{RHL}}=\lim _{x \rightarrow 1^{+}}(3-\mathrm{x})=3-1=2$ \\$\because \mathrm{LHL}=\mathrm{RHL}=2
and f(1)=3-x=3-1=2
\therefore f(x)$ is continuous at $x=1
Hence, condition 1 is satisfied.
Condition 2:
Now, we have to check   f(x)  is differentiable
f(x)=\left\{\begin{array}{l}x^{2}+1, \text { if } 0 \leq x \leq 1 \\ 3-x, \text { if } 1 \leq x \leq 2\end{array}\right.
On differentiating with respect to   x,  we get
\Rightarrow f^{\prime}(x)=\left\{\begin{array}{c}2 x+0, \text { if } 0<x<1 \\ 0-1, \text { if } 1<x<2\end{array}\right. or

\mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{l} 2 \mathrm{x}, \text { if } 0<\mathrm{x}<1 \\ -1, \text { if } 1<\mathrm{x}<2 \end{array}\right.

Now, let us consider the differentiability of f(x) at x = 1

LHD ⇒ f(x) = 2x = 2(1) = 2

RHD ⇒ f(x) = -1 = -1

LHD ≠ RHD

∴ f(x) is not differentiable at x = 1

Thus, Rolle’s theorem is not applicable to the given function.

Posted by

infoexpert22

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