PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Find the Maximum and Minimum Values of the Function F(X) = 4 X + 2 + X . - PUC Karnataka Science Class 12 - Mathematics

#### Question

Find the maximum and minimum values of the function f(x) = $\frac{4}{x + 2} + x .$

#### Solution

$\text { Given }: f\left( x \right) = \frac{4}{x + 2} + x$

$\Rightarrow f'\left( x \right) = - \frac{4}{\left( x + 2 \right)^2} + 1$

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

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

$\Rightarrow - \frac{4}{\left( x + 2 \right)^2} + 1 = 0$

$\Rightarrow - \frac{4}{\left( x + 2 \right)^2} = - 1$

$\Rightarrow \left( x + 2 \right)^2 = 4$

$\Rightarrow x + 2 = \pm 2$

$\Rightarrow x = 0 \text { and } - 4$

$\text { Thus, x = 0 and x = - 4 are the possible points of local maxima or local minima } .$

$\text { Now, }$

$f''\left( x \right) = \frac{8}{\left( x + 2 \right)^3}$

$\text { At x } = 0:$

$f''\left( 0 \right) = \frac{8}{\left( 2 \right)^3} = 1 > 0$

$\text { So, x = 0 is a point of local minimum } .$

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

$f\left( 0 \right) = \frac{4}{0 + 2} + 0 = 2$

$\text { At x } = - 4:$

$f''\left( - 4 \right) = \frac{8}{\left( - 4 \right)^3} = \frac{- 1}{8} < 0$

$\text { So, x = - 4 is a point of local minimum } .$

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

$f\left( - 4 \right) = \frac{4}{- 4 + 2} - 4 = - 6$

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Solution Find the Maximum and Minimum Values of the Function F(X) = 4 X + 2 + X . Concept: Graph of Maxima and Minima.
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