Question

The value of c in Rolle's theorem for the function f(x) = x³ - 3x in the interval [0, $\sqrt{3}$] is _______________.

Answer 1

The given function is f(x) = x³ − 3x.

f(x) is a polynomial function. We know that a polynomial function is everywhere continuous and differentiable.

So, f(x) is continuous on $\left[0,\sqrt{3}\right]$ and differentiable on $\left(0,\sqrt{3}\right)$.

Also, f(0) = 0 and $f\left(\sqrt{3}\right)={\left(\sqrt{3}\right)}^3-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0$

$\therefore f\left(0\right)=f\left(\sqrt{3}\right)$

Thus, all the conditions of Rolle's theorem are satisfied.

So, there exist a real number c ∈ $\left(0,\sqrt{3}\right)$ such that $f\text{'}\left(c\right)=0$.

f(x) = x³ − 3x

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

$\therefore f\text{'}\left(c\right)=0$

$\Rightarrow 3c^2-3=0$

$\Rightarrow c^2=1$

$\Rightarrow c=\pm 1$

Thus, c = 1 ∈ $\left(0,\sqrt{3}\right)$ such that $f\text{'}\left(c\right)=0$.

Hence, the value of c is 1.


The value of c in Rolle's theorem for the function f(x) = x³ − 3x in the interval $\left[0,\sqrt{3}\right]$ is ___1___.