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# F(X) = X + a 2 X , a > 0 , , X ≠ 0 . - Mathematics

#### Question

f(x) = $x + \frac{a2}{x}, a > 0,$ , x ≠ 0 .

#### Solution

$\text { Given }: f\left( x \right) = x + \frac{a^2}{x}$

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

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

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

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

$\Rightarrow x^2 = a^2$

$\Rightarrow x = \pm a$

$\text { Thus, x = a and x = - a are the possible points of local maxima or local minima }.$

$\text { Now,}$

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

$\text { At x = a }:$

$f''\left( a \right) = \frac{a^2}{\left( a \right)^3} = \frac{1}{a} > 0$

$\text { So, x = a is the point of local minimum } .$

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

$f\left( a \right) = x + \frac{a^2}{x} = a + a = 2a$

$At x = - a:$

$f''\left( a \right) = \frac{a^2}{\left( - a \right)^3} = - \frac{1}{a} < 0$

$\text { So, x = - a is the point of local maximum }.$

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

$f\left( - a \right) = x + \frac{a^2}{x} = - a - a = - 2a$

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#### APPEARS IN

RD Sharma Solution for Mathematics for Class 12 (Set of 2 Volume) (2018 (Latest))
Chapter 18: Maxima and Minima
Ex. 18.3 | Q: 1.1 | Page no. 31

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F(X) = X + a 2 X , a > 0 , , X ≠ 0 . Concept: Graph of Maxima and Minima.