Question

The maximum value of f(x) = \[\frac{x}{4 - x + x^2}\] on [ \[-\] 1, 1] is _______________ .

Options

• \[ \frac{1}{4}\]

• \[- \frac{1}{3}\]

• \[\frac{1}{6}\]

• \[\frac{1}{5}\]

Answer 1

\[\text { Given: } f\left( x \right) = \frac{x}{4 - x + x^2}\]

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

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

\[f'\left( x \right) = 0\]

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

\[ \Rightarrow 4 - x + x^2 - x\left( - 1 + 2x \right) = 0\]

\[ \Rightarrow 4 - x + x^2 + x - 2 x^2 = 0\]

\[ \Rightarrow x^2 = 4\]

\[ \Rightarrow x =\pm 2 \not\in \left( - 1, 1 \right)\]

\[\text { So,} \]

\[f\left( - 1 \right) = \frac{- 1}{4 - \left( - 1 \right) + \left( - 1 \right)^2} = \frac{- 1}{6}\]

\[f\left( 1 \right) = \frac{1}{4 - 1 + 1^2} = \frac{1}{4}\]

\[\text { Hence, the maximum value is } \frac{1}{4} . \]