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

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(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

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

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(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 $\rule{1cm}{0.15mm}$

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