Let $g (x) = x^{3} - 6 x^{2} + 9 x + 2$ on $[0, 3]$ . Show that $g$ satisfies Rolle's theorem and find all $c \in (0, 3)$ such that $g^{'} (c) = 0$ .

hard 4 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c = 1 and c = 2

The correct answer is $c = 1 and c = 2$ — Hence the value(s) of $c$ in $(0, 3)$ with $g^{'} (c) = 0$ is $c = 1$ .

Step-by-step solution

Check continuity on $[0, 3]$ – $g$ is a polynomial, hence continuous everywhere.
Check differentiability on $(0, 3)$ – polynomials are differentiable everywhere.
Evaluate end points: $g (0) = 2$ , $g (3) = 27 - 54 + 27 + 2 = 2$ . Since $g (0) = g (3)$ , all three hypotheses of Rolle's theorem hold.
Compute derivative: $g^{'} (x) = 3 x^{2} - 12 x + 9$ .
Set $g^{'} (c) = 0$ : $3 c^{2} - 12 c + 9 = 0 \Rightarrow c^{2} - 4 c + 3 = 0 \Rightarrow (c - 1) (c - 3) = 0$ . Wait, factor correctly: $c^{2} - 4 c + 3 = 0$ gives $c = 1$ or $c = 3$ . But $c$ must lie in $(0, 3)$ , so $c = 1$ only. Re‑check derivative: actually $g^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ . Thus zeros are $x = 1$ and $x = 3$ . $x = 3$ is an endpoint, not interior, so the only interior point is $c = 1$ .
Hence the value(s) of $c$ in $(0, 3)$ with $g^{'} (c) = 0$ is $c = 1$ .

Common mistake: Forgetting to verify

g (0) = g (3)

or including endpoint as interior solution

Concept tested: Rolle's theorem and finding stationary points