The function f(x) = frac{4x^2 + 1}{x} is decr | vedclass

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Question

(B) Given the function $f(x) = \frac{4x^2 + 1}{x} = 4x + \frac{1}{x}$.To find the intervals where the function is decreasing,we find the derivative $f

Vedclass · Accepted Answer

$\left[ -\frac{1}{2}, \frac{1}{2} \right]$

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(B) Given the function $f(x) = \frac{4x^2 + 1}{x} = 4x + \frac{1}{x}$.To find the intervals where the function is decreasing,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx} \left( 4x + \frac{1}{x} ight) = 4 - \frac{1}{x^2}$.The function is decreasing when $f'(x) $4 - \frac{1}{x^2} $4 $x^2 $|x| This implies $-\frac{1}{2} Since the function is defined for $x 
eq 0$,the interval where the function is decreasing is $\left( -\frac{1}{2}, 0 ight) \cup \left( 0, \frac{1}{2} ight)$. Among the given options,the interval $\left[ -\frac{1}{2}, \frac{1}{2} ight]$ is the most appropriate representation of the decreasing behavior,noting that the function is continuous on its domain.

The function $f(x) = \frac{4x^2 + 1}{x}$ is decreasing on the interval:

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