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

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\text { Given: } f\left( x \right) = 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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