The value of ‘c’ by Lagrange’s theorem,if f(x)=2+x3x≤13xx>1 , on [-1,2]

f1(c)=3c2=f(2)−f(−1)2−(−1)=6−13=53

The value of ‘c’ by Lagrange’s theorem,if f(x)=2+x3x≤13xx>1 , on [-1,2]

f1(c)=3c2=f(2)−f(−1)2−(−1)=6−13=53

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The value of ‘c’ by Lagrange’s theorem,if $f (x) = \{\begin{cases} 2 + x^{3} & x \leq 1 \\ 3 x & x > 1 \end{cases}$ , on [-1,2]

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$\frac{\pm \sqrt{5}}{3}$

$\frac{\pm \sqrt{3}}{2}$

$\frac{\pm \sqrt{2}}{5}$

$\frac{\pm 3}{\sqrt{5}}$

answer is A.

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$f^{1} (c) = 3 c^{2} = \frac{f (2) - f (- 1)}{2 - (- 1)} = \frac{6 - 1}{3} = \frac{5}{3}$

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The value of ‘c’ by Lagrange’s theorem,if f(x)=2+x3x≤13xx>1 , on [-1,2]

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The value of ‘c’ by Lagrange’s theorem,if $f (x) = \{\begin{cases} 2 + x^{3} & x \leq 1 \\ 3 x & x > 1 \end{cases}$ , on [-1,2]