The maximum value of f(x) = frac{1}{4x^2 + 2x | vedclass

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(A) Given $f(x) = \frac{1}{4x^2 + 2x + 1}$.To find the maximum value,we first find the derivative $f'(x)$:$f'(x) = -\frac{1}{(4x^2 + 2x + 1)^2} \cdot

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$4/3$

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(A) Given $f(x) = \frac{1}{4x^2 + 2x + 1}$.To find the maximum value,we first find the derivative $f'(x)$:$f'(x) = -\frac{1}{(4x^2 + 2x + 1)^2} \cdot (8x + 2) = -\frac{2(4x + 1)}{(4x^2 + 2x + 1)^2}$.Setting $f'(x) = 0$ gives $4x + 1 = 0$,so $x = -1/4$.Since the denominator $4x^2 + 2x + 1$ is always positive (as its discriminant $D = 2^2 - 4(4)(1) = 4 - 16 = -12 The minimum value of the quadratic $g(x) = 4x^2 + 2x + 1$ occurs at $x = -b/(2a) = -2/(2 \cdot 4) = -1/4$.The minimum value of the denominator is $g(-1/4) = 4(1/16) + 2(-1/4) + 1 = 1/4 - 1/2 + 1 = 3/4$.Therefore,the maximum value of $f(x)$ is $1 / (3/4) = 4/3$.

The maximum value of $f(x) = \frac{1}{4x^2 + 2x + 1}$ is .....

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