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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Vedclass · Answer

(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:

Similar Questions

If $f(x) = x^3 - x^2 + 100x + 1001$,then:

The number of real solutions of $x^{7}+5x^{3}+3x+1=0$ is equal to............

The function $f(x) = x^x$ for $x > 0$ is strictly increasing at

What is the nature of the function $f(x) = \sin x$ at $x = 2\pi / 3$?

In which interval is the function $f(x) = \log x - \frac{2x}{2 + x}$ an increasing function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module