Examine the applicable of MVT for all three functions.

$f (x) = 1 - x^{2} f o r x ϵ [1, 2]$

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Solution

Here, $f (x) = 1 - x^{2}$ , which is a polynomial function.

Since, a polynomial function is continuous and derivable at all $x ϵ R$ , therefore

(a) f is continuous in [1, 2]

(b) f is derivable in (1, 2) $\Rightarrow$ f satisfies conditions of MVT in [1, 2]

Hence, there exists atleast one real c $ϵ$ [1, 2] such that

$f^{'} (c) = \frac{f (2) - f (1)}{2 - 1} (∴ f^{'} (c) = - \frac{f (b) - f (a)}{b - a}) \Rightarrow - 2 c = \frac{(1 - 2^{2}) - (1 - 1^{2})}{2 - 1} (∵ f (x) = 1 - x^{2}, ∴ f^{'} (x) = - 2 x) \Rightarrow - 2 c = - 3 \Rightarrow c = \frac{3}{2} ϵ (1, 2) ∴ M V T i s v e r i f i e d f o r t h e g i v e n f u n c t i o n i n t h e g i v e n i n t e r v a l .$

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Examine the applicable of MVT for all three functions. f(x)=1−x2 for xϵ[1, 2]

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