The function f(x) = tan x - 4x is strictly de | vedclass

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

Question

(A) function $f(x)$ is strictly decreasing on an interval if its derivative $f'(x) < 0$ for all $x$ in that interval.Given $f(x) = 	an x - 4x$.Taking

Vedclass · Accepted Answer

$(-\frac{\pi}{3}, \frac{\pi}{3})$

Vedclass · Answer

(A) function $f(x)$ is strictly decreasing on an interval if its derivative $f'(x) Given $f(x) = 	an x - 4x$.Taking the derivative with respect to $x$, we get $f'(x) = \sec^2 x - 4$.For the function to be strictly decreasing, we require $f'(x) $\sec^2 x - 4 $\sec^2 x Taking the square root on both sides, we get $|\sec x| Since $|\sec x| = 1/|\cos x|$, this implies $1/|\cos x|  1/2$.This inequality holds when $-\frac{\pi}{3} < x < \frac{\pi}{3}$.

The function $f(x) = \tan x - 4x$ is strictly decreasing on $\rule{1cm}{0.15mm}$

Similar Questions

For what value of $\lambda$ is the function $f(x) = \lambda x + \cos x$ strictly increasing?

If $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2), \forall x \in R$,where $f''(x) > 0, \forall x \in R$. Then $g(x)$ is increasing in the interval

For all real values of $x$,the increasing function $f(x)$ is

What type of function is $y = x^4$?

For what values of $x$ is the function $f(x) = x + \frac{1}{x}, x \neq 0$ strictly increasing?

Vedclass Test Series

Exam Paper Generator

Online Exam Module