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

Where does $f(x) = x + \sqrt{1 - x}, 0 < x < 1$ decrease?

If $f(x) = \frac{1}{x + 1} - \log(1 + x)$,where $x > 0$,then what type of function is $f$?

For what values of $a$ is the function $f(x) = x^{2} + ax + 1$ increasing on the interval $[1, 2]$?

For what values of $K$ is the function $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$ strictly increasing for all $x \in \mathbb{R}$?

Find the intervals in which the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing or strictly decreasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module