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प्रश्न
For the function f(x) = \[x + \frac{1}{x}\]
पर्याय
x = 1 is a point of maximum
x = \[-\] 1 is a point of minimum
maximum value > minimum value
maximum value < minimum value
उत्तर
\[\text { maximum value < minimum value}\]
\[\text { Given:} f\left( x \right) = x + \frac{1}{x}\]
\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]
\[\text { For a local maxima or a local minima, we must have} \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]
\[ \Rightarrow x^2 - 1 = 0\]
\[ \Rightarrow x^2 = 1\]
\[ \Rightarrow x = \pm 1\]
\[\text { Now }, \]
\[f''\left( x \right) = \frac{2}{x^3}\]
\[ \Rightarrow f''\left( 1 \right) = \frac{2}{1} = 2 > 0\]
\[\text { So, x = 1 is a local minima.}\]
\[\text { Also }, \]
\[f''\left( - 1 \right) = - 2 < 0\]
\[\text {So, x = - 1 is a localmaxima }.\]
\[\text { The local minimum value is given by }\]
\[f\left( 1 \right) = 2\]
\[\text { The local maximum value is given by }\]
\[f\left( - 1 \right) = - 2\]
\[ \therefore \text { Maximum value < Minimum value }\]
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