In which of the following functions is Rolle' | vedclass

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(D) For Rolle's theorem to be applicable,the function must be continuous on $[a, b]$,differentiable on $(a, b)$,and $f(a) = f(b)$.$(A)$ The function i

Vedclass · Accepted Answer

$f(x) = \begin{cases} \frac{x^3 - 2x^2 - 5x + 6}{x - 1}, & x 
e 1 \ -6, & x = 1 \end{cases}$ on $[-2, 3]$

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(D) For Rolle's theorem to be applicable,the function must be continuous on $[a, b]$,differentiable on $(a, b)$,and $f(a) = f(b)$.$(A)$ The function is discontinuous at $x = 1$,so Rolle's theorem is not applicable.$(B)$ The function is not continuous at $x = 0$ because $\lim_{x 	o 0^-} \frac{\sin x}{x} = 1 
e f(0) = 0$,so Rolle's theorem is not applicable.$(C)$ The function is discontinuous at $x = 1$,which lies in the interval $[-2, 3]$,so Rolle's theorem is not applicable.$(D)$ For $x 
e 1$,$f(x) = \frac{(x-1)(x^2 - x - 6)}{x - 1} = x^2 - x - 6$. At $x = 1$,$f(1) = -6$. Since $\lim_{x 	o 1} (x^2 - x - 6) = 1 - 1 - 6 = -6$,the function is continuous at $x = 1$. Thus,$f(x) = x^2 - x - 6$ for all $x \in [-2, 3]$. Also,$f(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0$ and $f(3) = 3^2 - 3 - 6 = 9 - 3 - 6 = 0$. Since $f(-2) = f(3) = 0$ and the function is a polynomial,it is continuous and differentiable. Hence,Rolle's theorem is applicable.

In which of the following functions is Rolle's theorem applicable?

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