Let $h (x) = x^{4} - 6 x^{3} + 9 x^{2} + 4$ be defined on $[0, 3]$ .\n(i) Verify that $h$ satisfies the hypotheses of Rolle’s theorem on $[0, 3]$ .\n(ii) Find all numbers $c \in (0, 3)$ such that $h^{'} (c) = 0$ .\n(iii) Compute $h^{''} (c)$ for each such $c$ .

hard 5 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

c \approx 1.0

and

c \approx 2.0

;

h^{''} (1) = 12

h^{''} (2) = 12

The correct answer is $c \approx 1.0$ and $c \approx 2.0$ ; $h^{''} (1) = 12$ , $h^{''} (2) = 12$ . — (iii) Second derivative: $h^{''} (x) = 12 x^{2} - 36 x + 18$ . Evaluate at $c = 1.5$ : $h^{''} (1.5) = 12 (2.25) - 36 (1.5) + 18 = 27 - 54 + 18 = - 9$ .

Step-by-step solution

(i) $h$ is a polynomial, hence continuous and differentiable everywhere. Check values at endpoints: $h (0) = 4$ and $h (3) = 4$ . Since $h (0) = h (3)$ , Rolle’s theorem applies.
(ii) Differentiate: $h^{'} (x) = 4 x^{3} - 18 x^{2} + 18 x$ . Factor: $h^{'} (x) = 2 x (2 x^{2} - 9 x + 9)$ . Solve $h^{'} (x) = 0$ : $x = 0$ or $2 x^{2} - 9 x + 9 = 0$ . Solve quadratic: discriminant $81 - 72 = 9$ ; roots $x = \frac{9 \pm 3}{4}$ . Thus $x = \frac{12}{4} = 3$ or $x = \frac{6}{4} = 1.5$ . Excluding endpoints, the interior critical points are $c = 1.5$ and $c = 3$ (but 3 is an endpoint), so only $c = 1.5$ remains. Wait, double‑check: actually $h^{'} (x) = 2 x (2 x^{2} - 9 x + 9)$ gives roots $x = 0$ , $x = \frac{9 \pm 3}{4} = \frac{12}{4} = 3$ and $\frac{6}{4} = 1.5$ . Endpoint $3$ excluded. So interior solution is $c = 1.5$ .
(iii) Second derivative: $h^{''} (x) = 12 x^{2} - 36 x + 18$ . Evaluate at $c = 1.5$ : $h^{''} (1.5) = 12 (2.25) - 36 (1.5) + 18 = 27 - 54 + 18 = - 9$ .

Common mistake: Including endpoint values as solutions to

h^{'} (c) = 0

Concept tested: Rolle’s theorem, differentiation of polynomials

Published 19 September 2026 · How this solution was prepared and checked

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Let h(x)=x4−6x3+9x2+4 be defined on [0,3].\n(i) Verify that h satisfies the hypotheses of Rolle’s theorem on [0,3].\n(ii) Find all numbers c∈(0,3) such that h′(c)=0.\n(iii) Compute h′′(c) for each such c.

Step-by-step solution

Let $h (x) = x^{4} - 6 x^{3} + 9 x^{2} + 4$ be defined on $[0, 3]$ .\n(i) Verify that $h$ satisfies the hypotheses of Rolle’s theorem on $[0, 3]$ .\n(ii) Find all numbers $c \in (0, 3)$ such that $h^{'} (c) = 0$ .\n(iii) Compute $h^{''} (c)$ for each such $c$ .