Question

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

Answer 1

Rolle’s Theorem is a special case of the mean-value theorem, which states that any real-valued differential equation whose values are equal at two different points must have a stationary point between those points. In general, it states that if a function is continuous on closed interval $\left[ {a,b} \right]$ and differential in open interval $\left( {a,b} \right)$ such that $f\left( a \right) = f\left( b \right),$ then $f'\left( x \right) = 0$ where $a \leqslant x \leqslant b$.
To check whether the given function is continuous in a range, we have to check whether the function is polynomial as we know polynomial functions are continuous everywhere in the range $\left( { - \infty ,\infty } \right)$.

Complete step by step answer:
The value of c must lie in the interval $\left[ { - \sqrt 3 ,0} \right]$ , and this is possible only when the given function is continuous in the closed interval $\left[ { - \sqrt 3 ,0} \right]$ and differentiable on the open interval $\left( { - \sqrt 3 ,0} \right)$ at $f'\left( c \right) = 0$.
Now, check whether the given function $f\left( x \right) = {x^3} - 3x$ is polynomial since there are no negative powers on $x$ in the cubic function; hence, it is a polynomial function. Now, check for the condition of Rolle’s Theorem to find the value of c.
For the first condition, as we know, the function is polynomial, so it will be continuous everywhere; hence it will be continuous in the range $\left[ { - \sqrt 3 ,0} \right]$.
Now for the second condition, where a polynomial function is differentiable everywhere in the range; hence the second condition is satisfied.
Now check the value of range for the function $\left[ { - \sqrt 3 ,0} \right]$,
At $x = - \sqrt 3$

$$f\left( x \right) = {x^3} - 3x \\\ f\left( { - \sqrt 3 } \right) = \left( {{{\left( { - \sqrt 3 } \right)}^3} - 3\left( { - \sqrt 3 } \right)} \right) \\\ = \left( { - 3\sqrt 3 + 3\sqrt 3 } \right) \\\ = 0 \\\$$

At $x = 0$

$$f\left( x \right) = {x^3} - 3x \\\ f\left( 0 \right) = \left( {{{\left( 0 \right)}^3} - 3\left( 0 \right)} \right) \\\ = 0 \\\$$

Hence, $f\left( { - \sqrt 3 } \right) = f\left( 0 \right)$
the Mean value theorem satisfies, and we can use Rolle’s Theorem as:
$f'\left( c \right) = 0$ for $c \in \left( { - \sqrt 3 ,0} \right)$
Therefore,

$$f\left( x \right) = {x^3} - 3x \\\ f'\left( x \right) = 3{x^2} - 3 \\\ f'\left( c \right) = 3{c^2} - 3 \\\$$

for Rolle’s theorem
$f'\left( c \right) = 0$

$$3{c^2} - 3 = 0 \\\ {c^2} = \dfrac{3}{3} \\\ c = \sqrt 1 \\\ = \pm 1 \\\$$

From $\pm 1$, $- 1$ lies in the interval $\left[ { - \sqrt 3 ,0} \right].$

Hence the value of $c = - 1$.

Note: To satisfy the condition for Mean Value Theorem, check whether the given function is a polynomial function. A polynomial function is a function that does not contain any negative integer of power x in the equation of a cubic, quadratic function, etc.