Question

The value of c in Rolle's theorem for the function f (x) = x³ − 3x in the interval [0, $\sqrt{3}$] is
(a) 1
(b) −1
(c) 3/2
(d) 1/3

Answer 1

(a) 1

The given function is $f\left(x\right)=x^3-3x$.
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0, $\sqrt{3}$] and derivable on (0, $\sqrt{3}$ ).

Differentiating the given function with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=3x^2-3 \\ \Rightarrow f\text{'}\left(c\right)=3c^2-3 \\ \therefore f\text{'}\left(c\right)=0 \\ \Rightarrow 3c^2-3=0 \\ \Rightarrow c^2=1 \\ \Rightarrow c=\pm 1 \end{gathered}$$

Thus, $c=1\in \left[0,\sqrt{3}\right]$ for which Rolle's theorem holds.

Hence, the required value of c is 1.