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

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$\frac{1}{6}$ \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^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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