If(x) = $\frac{1}{4}$x^2 + 2x + 1 then its maximum value is _________________ .

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KollegeApply · Accepted Answer

$\frac{4}{3}$ \text { Maximum value of }$\frac{1}{4}$ x^2 + 2x + 1= \text { Minimum value of }4 x^2 + 2x + 1 \text{ Now, } f\left( x \right) = 4 x^2 + 2x + 1 \Rightarrow f'\left( x \right) = 8x + 2 \text { For a local maxima or a local minima, we must have } f'\left( x \right) = 0 \Rightarrow 8x + 2 = 0 \Rightarrow 8x = - 2 \Rightarrow x = - $\frac{1}{4}$ \text { Now, } f''\left( x \right) = 8 \Rightarrow f''\left( 1 \right) = 8 > 0 \text { So, x } = - $\frac{1}{4}$ \text { is a local minima } . \text { Thus },$\frac{1}{4}$ x^2 + 2x + 1\text { is maximum at x } = - $\frac{1}{4}$ . \Rightarrow \text { Maximum value of } $\frac{1}{4}$ x^2 + 2x + 1 = \frac{1}{4 \left( - $\frac{1}{4}$ \right)^2 + 2\left( - $\frac{1}{4}$ \right) + 1} = 1/4/16 - $\frac{1}{2}$ + 1 = $\frac{16}{12}$ = $\frac{4}{3}$

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