In which of the following function is rolle’s theorem applicable?

Easy

A

$f (x) = {\begin{matrix} x, 0 \leq x \leq 1 \\ 0, x = 1 \end{matrix} o n [0, 1]$
B

$f (x) = {\begin{matrix} \frac{\sin x}{x}, - π \leq x \leq o \\ 0, x = 0 \end{matrix} o n [- π, 0]$
C

$f (x) = \frac{x^{2} - x - 6}{x - 1} o n [- 2.3]$
D

$f (x) = {\begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, i f x \neq 1, \\ - 6, i f x = 1 \end{matrix}$

Solution

A) Discontinuous at x=1 $\Rightarrow$ not applicable.
B) $F (x)$ is not continuous (jump discontinuity) at x=0.

C) Discontinuity (missing point) at x=1 $\Rightarrow$ not applicable.

D) notice that $x^{3} - 2 x^{2} - 5 x + 6 = (x - 1) (x^{2} - x - 6) .$

hence, $f (x) = x^{2} - x - 6 i f x \neq 1 a n d f (1) = - 6.$

$\Rightarrow f$ is continuous at x=1. So $f (x) = x^{2} - x - 6$ throughout the interval $[- 2, 3] .$

Also, note that $f (- 2) = f (3) = 0.$ hence, Rolle’s theorem applies. $f^{'} (x) = 2 x - 1.$

Setting $f^{'} (x) = 0,$ we obtain $x = \frac{1}{2}$ which lies between –2 and 3.

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