Verify that $p (x) = x^{3} - 3 x$ satisfies the hypotheses of Rolle’s theorem on $[0, 3]$ and find all numbers $c$ in $(0, 3)$ such that $p^{'} (c) = 0$ .

medium 3 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c = 1

The correct answer is c = 1 — Hence the required $c$ is $1$ .

Step-by-step solution

Check continuity: $p (x)$ is a polynomial, hence continuous on all real numbers, including $[0, 3]$ .
Check differentiability: $p (x)$ is a polynomial, hence differentiable everywhere, including $(0, 3)$ .
Evaluate the function at the endpoints: $p (0) = 0$ and $p (3) = (3)^{3} - 33 = 33 - 33 = 0$ . So $p (0) = p (3)$ .
By Rolle’s theorem, there exists at least one $c \in (0, 3)$ with $p^{'} (c) = 0$ .
Find the derivative: $p^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1)$ .
Set $p^{'} (c) = 0$ : $3 (c^{2} - 1) = 0 \Rightarrow c^{2} - 1 = 0 \Rightarrow c = \pm 1$ .
Only $c = 1$ lies in $(0, 3)$ .
Hence the required $c$ is $1$ .

Common mistake: Missing the condition

p (0) = p (3)

or selecting

c = - 1

Concept tested: Rolle’s theorem application