Question

4 multiple answer 1 point select all statements that are true about the function, ( f(x)=\tan x ). the range of ( f(x)=\tan x ) is ( (-infty, infty) ). the graph of ( f(x)=\tan x ) is always increasing. the tangent function is negative in every quadrant expect for quadrant i. the period of the graph of ( f(x)=\tan x ) is ( 2 pi ).

Explanation:

Step1: Analyze the range of \(y = \tan x\)

The tangent function \(y=\tan x=\frac{\sin x}{\cos x}\). As \(x\) approaches \(\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) from the left, \(\tan x\to+\infty\), and as \(x\) approaches \(\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) from the right, \(\tan x\to-\infty\). So the range of \(y = \tan x\) is \((-\infty,\infty)\).

Step2: Analyze the monotonicity of \(y=\tan x\)

The derivative of \(y = \tan x\) is \(y'=\sec^{2}x=\frac{1}{\cos^{2}x}>0\) for all \(x
eq\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\). In each interval \((-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi),k\in\mathbb{Z}\), the function \(y = \tan x\) is increasing. But we cannot say "the graph of \(y=\tan x\) is always increasing" in the entire domain (because of the discontinuities at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\)).

Step3: Analyze the sign of \(y=\tan x\)

Using the formula \(\tan x=\frac{\sin x}{\cos x}\). In Quadrant I (\(0 < x<\frac{\pi}{2}\)), \(\sin x>0\) and \(\cos x>0\), so \(\tan x>0\). In Quadrant II (\(\frac{\pi}{2}0\) and \(\cos x < 0\), so \(\tan x<0\). In Quadrant III (\(\pi0\). In Quadrant IV (\(\frac{3\pi}{2}0\), so \(\tan x<0\).

Step4: Analyze the period of \(y=\tan x\)

The period of \(y = \tan x\) is \(\pi\) (since \(\tan(x + \pi)=\frac{\sin(x+\pi)}{\cos(x+\pi)}=\frac{-\sin x}{-\cos x}=\tan x\))

Answer:

The first statement (The range of \(f(x)=\tan x\) is \((-\infty,\infty)\)) is True.