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Question
The maximum value of f(x) = \[\frac{x}{4 + x + x^2}\] on [ \[-\] 1,1] is ___________________ .
Options
\[- \frac{1}{4}\]
\[- \frac{1}{3}\]
\[\frac{1}{6}\]
\[\frac{1}{5}\]
Solution
\[\frac{1}{6}\]
\[\text { Given }: f\left( x \right) = \frac{x}{4 + x + x^2}\]
\[ \Rightarrow f'\left( x \right) = \frac{4 + x + x^2 - x\left( 1 + 2x \right)}{\left( 4 + x + x^2 \right)^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{4 + x + x^2 - x\left( 1 + 2x \right)}{\left( 4 + x + x^2 \right)^2} = 0\]
\[ \Rightarrow 4 + x + x^2 - x\left( 1 + 2x \right) = 0\]
\[ \Rightarrow 4 - x^2 = 0\]
\[ \Rightarrow x = \pm 2 \not\in \left[ - 1, 1 \right]\]
\[\text { The values of } f\left( x \right) \text { at extreme points are given by }\]
\[f\left( 1 \right) = \frac{1}{4 + 1 + 1^2} = \frac{1}{6}\]
\[f\left( - 1 \right) = \frac{- 1}{4 - 1 + \left( - 1 \right)^2} = \frac{- 1}{4}\]
\[\text{Thus,}\frac{1}{6}\text{ is the maximum value }\] .
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