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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
\[\text { Given: } f\left( x \right) = \frac{x}{4 - x + x^2}\]
\[ \Rightarrow f'\left( x \right) = \frac{4 - x + x^2 - x\left( - 1 + 2x \right)}{\left( 4 - x + x^2 \right)^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{4 - x + x^2 - x\left( - 1 + 2x \right)}{\left( 4 - x + x^2 \right)^2} = 0\]
\[ \Rightarrow 4 - x + x^2 - x\left( - 1 + 2x \right) = 0\]
\[ \Rightarrow 4 - x + x^2 + x - 2 x^2 = 0\]
\[ \Rightarrow x^2 = 4\]
\[ \Rightarrow x =\pm 2 \not\in \left( - 1, 1 \right)\]
\[\text { So,} \]
\[f\left( - 1 \right) = \frac{- 1}{4 - \left( - 1 \right) + \left( - 1 \right)^2} = \frac{- 1}{6}\]
\[f\left( 1 \right) = \frac{1}{4 - 1 + 1^2} = \frac{1}{4}\]
\[\text { Hence, the maximum value is } \frac{1}{4} . \]
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