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  • The value of c in Rolle’s theorem for the function f(x) = x^3 - 3x in the interval [0,root3] A. 1

The value of c in Rolle’s theorem for the function f(x) = x^3 - 3x in the interval [0,\sqrt 3] is
A. 1

B.-1

C. \frac{3}{2}

D.\frac{1}{3}

Answers (1)

A)

\\ \text{Rolle's Theorem states that, Let} \ f_{i} \ [a, b] \rightarrow R \text{be continuous on} \ [a, b] \text{and differentiable on} \ (a, b), \text{such that} \ f(a)=f(b), \text{where a and b are some real numbers.} \\ \text{Then there exists some c in (a, b) such that} \ f^{\prime}(c)=0 \\ We \ have, f(x)=x^{3}-3 x \ since, \\ f(x) \text{is a polynomial function it is continuous on} \ [0, \sqrt{3}] \ {\text {and }} \text{differentiable on} (0, \sqrt{3})\\ f(0)=f(\sqrt{3})=0 \\f^{\prime}(c)=0 -3 c^{2}-3=0\left[\because f^{\prime}(x)=3 x^{2}-3\right] \\\Rightarrow \mathrm{c}^{2}=1 \\\Rightarrow c=\pm 1 \\c=1 \in (0, \sqrt{3})

 

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