Find the value of c in Rolle’s theorem for the function in

Find the value of c in Rolle’s theorem for the function $f\left ( x \right )= x^{3}-3x$ in $\left [ -\sqrt{3},0 \right ]\cdot$

Answers (1)

we know that the polynamial function $f\left ( x \right )= x^{2}-3x$ is every where contineous and differentiable
so, $f\left ( x \right )$ is contineous on $\left [ -\sqrt{3},0 \right ]$ Also $f\left ( x \right )$ is differentiable on $\left [ -\sqrt{3},0 \right ]$
Now $f\left ( -\sqrt{3} \right )= \left ( -\sqrt{3} \right )^{3}-3\left ( -\sqrt{3} \right )$
$= -3\sqrt{3}+3\sqrt{3}= 0$
and $f\left ( 0 \right )= \left ( 0 \right )^{3}-3\left ( 0 \right )= 0$
$\therefore f\left ( -\sqrt{3} \right )= f\left ( 0 \right )$
Thus, all the three condition of Rolle's theorem are satisfied
Now, there must exist $c\, \epsilon \, \left ( -\sqrt{3},0 \right )$ such that ${f}'\left ( c \right )= 0$
Now, ${f}'\left ( x \right )= 3x^{2}-3$ then ${f}'\left ( c \right )= 0$
$\Rightarrow 3c^{2}-3= 0\; \; \Rightarrow 3\left ( c^{2}-1 \right )= 0$
$\Rightarrow c= \pm 1$
$c\neq 1\, as\, 1\, \not{\epsilon }\left ( -\sqrt{3},0 \right )$
$\therefore c= -1\not{\epsilon }\left ( -\sqrt{3},0 \right )$
Thus, required value of c is -1