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F ( X ) = 2 X − 2 X 2 , X > 0 - Mathematics

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प्रश्न

`f(x) = 2/x - 2/x^2,  x>0`

बेरीज

उत्तर

\[\text { Given }: f\left( x \right) = \frac{2}{x} - \frac{2}{x^2} = 2 x^{- 1} - 2 x^{- 2} \]

\[ \Rightarrow f'\left( x \right) = - 2 x^{- 2} + 4 x^{- 3} = \frac{4}{x^3} - \frac{2}{x^2}\]

\[\text { For the local maxima or minima, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow \frac{4}{x^3} - \frac{2}{x^2} = 0\]

\[ \Rightarrow 4 - 2x = 0\]

\[ \Rightarrow x = 2\]

\[\text { Thus, x = 2 is the possible point of local maxima or local minima }. \]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{- 12}{x^4} + \frac{4}{x^3}\]

\[\text { At }x = 2: \]

\[ f''\left( 2 \right) = \frac{- 12}{16} + \frac{4}{8} = \frac{- 12 + 8}{16} = \frac{- 1}{4} < 0\]

\[\text { So, x = 2 is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( 2 \right) = \frac{2}{2} - \frac{2}{2^2} = 1 - \frac{1}{2} = \frac{1}{2}\]

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पाठ 18: Maxima and Minima - Exercise 18.3 [पृष्ठ ३१]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 18 Maxima and Minima
Exercise 18.3 | Q 1.04 | पृष्ठ ३१

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