If f(x) = frac{1}{4x^2 + 2x + 1},then its max | vedclass

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(A) To find the maximum value of $f(x) = \frac{1}{4x^2 + 2x + 1}$,we first find the minimum value of the denominator $g(x) = 4x^2 + 2x + 1$.Since $g(x

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$4/3$

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(A) To find the maximum value of $f(x) = \frac{1}{4x^2 + 2x + 1}$,we first find the minimum value of the denominator $g(x) = 4x^2 + 2x + 1$.Since $g(x)$ is a quadratic expression of the form $ax^2 + bx + c$ with $a = 4 > 0$,its minimum value occurs at $x = -\frac{b}{2a} = -\frac{2}{2(4)} = -\frac{1}{4}$.The minimum value of the denominator is $g(-\frac{1}{4}) = 4(-\frac{1}{4})^2 + 2(-\frac{1}{4}) + 1 = 4(\frac{1}{16}) - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1-2+4}{4} = \frac{3}{4}$.Since $f(x) = \frac{1}{g(x)}$,the maximum value of $f(x)$ is the reciprocal of the minimum value of $g(x)$.Therefore,the maximum value is $\frac{1}{3/4} = \frac{4}{3}$.

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

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