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