PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for For the Function F(X) = X + 1 X (A) X = 1 is a Point of Maximum (B) X = − 1 is a Point of Minimum (C) Maximum Value > Minimum Value (D) Maximum Value< Minimum Value - PUC Karnataka Science Class 12 - Mathematics

#### Question

For the function f(x) = $x + \frac{1}{x}$

(a) x = 1 is a point of maximum
(b) x = $-$ 1 is a point of minimum
(c) maximum value > minimum value
(d) maximum value< minimum value

#### Solution

$(d) \text { maximum value < minimum value}$

$\text { Given:} f\left( x \right) = x + \frac{1}{x}$

$\Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}$

$\text { For a local maxima or a local minima, we must have}$

$f'\left( x \right) = 0$

$\Rightarrow 1 - \frac{1}{x^2} = 0$

$\Rightarrow x^2 - 1 = 0$

$\Rightarrow x^2 = 1$

$\Rightarrow x = \pm 1$

$\text { Now },$

$f''\left( x \right) = \frac{2}{x^3}$

$\Rightarrow f''\left( 1 \right) = \frac{2}{1} = 2 > 0$

$\text { So, x = 1 is a local minima.}$

$\text { Also },$

$f''\left( - 1 \right) = - 2 < 0$

$\text {So, x = - 1 is a localmaxima }.$

$\text { The local minimum value is given by }$

$f\left( 1 \right) = 2$

$\text { The local maximum value is given by }$

$f\left( - 1 \right) = - 2$

$\therefore \text { Maximum value < Minimum value }$

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Solution For the Function F(X) = X + 1 X (A) X = 1 is a Point of Maximum (B) X = − 1 is a Point of Minimum (C) Maximum Value > Minimum Value (D) Maximum Value< Minimum Value Concept: Graph of Maxima and Minima.
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