which of the following represents $y = \\cot x$ ?

which of the following represents $y = \\cot x$ ?

which of the following represents $y = \\cot x$ ?

Answer

Explanation:

Step1: Recall properties of ( y = \cot x )

The cotangent function ( y=\cot x=\frac{\cos x}{\sin x} ) has vertical asymptotes where ( \sin x = 0 ), i.e., at ( x = n\pi ) for integer ( n ). The period of ( \cot x ) is ( \pi ), and in the interval ( (0, \pi) ), ( \cot x ) is negative (since ( \cos x ) is positive and ( \sin x ) is positive in ( (0, \pi/2) ), but ( \cos x ) is negative and ( \sin x ) is positive in ( (\pi/2, \pi) )? Wait, no: ( \cos x ) in ( (0, \pi) ): ( \cos x ) is positive in ( (0, \pi/2) ), negative in ( (\pi/2, \pi) ); ( \sin x ) is positive in ( (0, \pi) ). So ( \cot x=\frac{\cos x}{\sin x} ) is positive in ( (0, \pi/2) ) (positive/positive) and negative in ( (\pi/2, \pi) ) (negative/positive). Wait, actually, let's check the sign in different quadrants. In the first quadrant (( 0 < x < \pi/2 )): ( \cos x>0 ), ( \sin x>0 ), so ( \cot x>0 ). In the second quadrant (( \pi/2 < x < \pi )): ( \cos x<0 ), ( \sin x>0 ), so ( \cot x<0 ). In the third quadrant (( \pi < x < 3\pi/2 )): ( \cos x<0 ), ( \sin x<0 ), so ( \cot x>0 ) (negative/negative). In the fourth quadrant (( 3\pi/2 < x < 2\pi )): ( \cos x>0 ), ( \sin x<0 ), so ( \cot x<0 ). Also, the graph of ( \cot x ) passes through ( ( \pi/4, 1) ), ( ( 3\pi/4, -1) ), etc.

Step2: Analyze the asymptotes and sign

The vertical asymptotes are at ( x = -\pi, 0, \pi ) (for the interval shown). Now, let's check the sign in intervals:

  • Between ( -\pi ) and ( 0 ): Let's take ( x = -\pi/2 ). ( \cot(-\pi/2)=\frac{\cos(-\pi/2)}{\sin(-\pi/2)}=\frac{0}{-1}=0 )? Wait, no: ( \cot(-\pi/2)=\frac{\cos(-\pi/2)}{\sin(-\pi/2)}=\frac{0}{-1}=0 )? Wait, ( \sin(-\pi/2)= -1 ), ( \cos(-\pi/2)=0 ), so ( \cot(-\pi/2)=0 ). Wait, actually, at ( x = -\pi/2 ), ( \cot x = 0 ). Let's check the interval ( (-\pi, 0) ): take ( x = -\pi/4 ). ( \cot(-\pi/4)=\frac{\cos(-\pi/4)}{\sin(-\pi/4)}=\frac{\sqrt{2}/2}{-\sqrt{2}/2}=-1 ). So in ( (-\pi, 0) ), let's see the quadrant: ( -\pi < x < 0 ) is equivalent to ( \pi < x + 2\pi < 2\pi ) (fourth quadrant), where ( \cos x>0 ), ( \sin x<0 ), so ( \cot x=\frac{\cos x}{\sin x}<0 ). Wait, maybe better to look at the standard graph: ( y = \cot x ) has vertical asymptotes at ( x = 0, \pi, -\pi ), etc. In the interval ( (0, \pi) ), the graph goes from ( +\infty ) (near ( x=0^+ )) to ( -\infty ) (near ( x=\pi^- )), passing through ( (\pi/2, 0) ). In the interval ( (-\pi, 0) ), it goes from ( -\infty ) (near ( x=-\pi^+ )) to ( +\infty ) (near ( x=0^- )), passing through ( (-\pi/2, 0) ).

Now let's analyze the graphs:

  • Top-left (blue): Asymptotes at ( -\pi, 0, \pi ). In ( (0, \pi) ), the graph is negative near ( 0 ) and positive near ( \pi )? Wait, no. Wait, the blue graph: between ( 0 ) and ( \pi ), the left part (near 0) is below the x-axis (negative), right part (near ( \pi )) is above? No, that doesn't match ( \cot x ).

  • Top-right (green): Asymptotes at ( -\pi, 0, \pi ). Passes through the origin? ( \cot 0 ) is undefined, so it can't pass through (0,0). So green is out.

  • Bottom-left (orange): Passes through the origin? ( \cot 0 ) is undefined, so orange is out (since it goes through (0,0)).

  • Bottom-right (red): Asymptotes at ( -\pi, 0, \pi ). Let's check the interval ( (0, \pi) ): near ( x=0^+ ), ( \cot x ) approaches ( +\infty ) (since ( \sin x ) is small positive, ( \cos x ) is near 1, so ( \cot x ) is large positive). Near ( x=\pi^- ), ( \sin x ) is small positive, ( \cos x ) is near -1, so ( \cot x ) is large negative. Wait, no: ( x ) near ( \pi^- ) (just less than ( \pi )), ( \sin x ) is positive (small), ( \cos x ) is negative (near -1), so ( \cot x = \cos x / \sin x ) is negative (large negative), so it goes from ( +\infty ) (near 0) to ( -\infty ) (near ( \pi )) in ( (0, \pi) ). In ( (-\pi, 0) ): near ( x=-\pi^+ ) (just more than ( -\pi )), ( \sin x ) is negative (small), ( \cos x ) is negative (near -1), so ( \cot x = \cos x / \sin x ) is positive (large positive). Near ( x=0^- ) (just less than 0), ( \sin x ) is negative (small), ( \cos x ) is positive (near 1), so ( \cot x = \cos x / \sin x ) is negative (large negative). Wait, maybe I got the sign wrong. Let's take ( x = \pi/4 ) (in ( (0, \pi/2) )): ( \cot(\pi/4)=1 ) (positive). ( x = 3\pi/4 ) (in ( (\pi/2, \pi) )): ( \cot(3\pi/4)= -1 ) (negative). So in ( (0, \pi) ), ( \cot x ) is positive in ( (0, \pi/2) ), negative in ( (\pi/2, \pi) ), passing through ( (\pi/2, 0) ). In ( (-\pi, 0) ), ( x = -\pi/4 ) (equivalent to ( 7\pi/4 ), fourth quadrant): ( \cot(-\pi/4)= -1 ) (negative). ( x = -3\pi/4 ) (equivalent to ( 5\pi/4 ), third quadrant): ( \cot(-3\pi/4)=1 ) (positive). So in ( (-\pi, 0) ), ( \cot x ) is negative in ( (-\pi/2, 0) ), positive in ( (-\pi, -\pi/2) ), passing through ( (-\pi/2, 0) ).

Now looking at the bottom-right (red) graph:

  • Near ( x=0^+ ): the graph is going up (towards ( +\infty ))? Wait, no, the red graph on the right side of ( x=0 ) (between 0 and ( \pi )): as ( x ) approaches ( 0^+ ), the graph is near the top (positive large), and as ( x ) approaches ( \pi^- ), it's near the bottom (negative large), passing through ( (\pi/2, 0) )? Wait, no, the red graph in ( (0, \pi) ): let's see the y-axis: at ( x=0 ), it's an asymptote. The red graph on the right of 0 (between 0 and ( \pi )): the curve is on the right side of the asymptote at 0, going from top (near 0) down to bottom (near ( \pi )), passing through the x-axis? Wait, maybe I made a mistake. Wait, the correct graph of ( y = \cot x ) has vertical asymptotes at ( x = n\pi ), and in each period ( (n\pi, (n+1)\pi) ), it decreases from ( +\infty ) to ( -\infty ), passing through ( x = n\pi + \pi/2 ) (where it crosses the x-axis). So in ( (0, \pi) ), it goes from ( +\infty ) (x→0⁺) to ( -\infty ) (x→π⁻), crossing the x-axis at ( \pi/2 ). In ( (-\pi, 0) ), it goes from ( -\infty ) (x→-π⁺) to ( +\infty ) (x→0⁻), crossing the x-axis at ( -\pi/2 ).

Now let's check the bottom-right (red) graph:

  • Asymptotes at ( -\pi, 0, \pi ). In ( (0, \pi) ): the graph is on the right side of the asymptote at 0 (x=0), going from top (positive) down to bottom (negative) as x approaches ( \pi ), which matches ( \cot x ) (since at ( x=\pi/2 ), it should cross the x-axis). Wait, does the red graph cross the x-axis? Let's see: the red graph in ( (0, \pi) ) seems to cross the x-axis (since there's a point where it's on the x-axis between 0 and ( \pi )). Similarly, in ( (-\pi, 0) ), it crosses the x-axis between ( -\pi ) and 0. Wait, maybe the bottom-right graph is the correct one? Wait, no, wait the top-left (blue) graph: let's check again.

Wait, maybe I confused the direction. Let's recall: ( y = \tan x ) has asymptotes at ( \pi/2 + n\pi ), and ( y = \cot x ) at ( n\pi ). The graph of ( \cot x ) is similar to ( -\tan(x - \pi/2) ). Alternatively, let's look at the standard graph: ( y = \cot x ) has vertical asymptotes at 0, ( \pi ), ( -\pi ), etc. In the interval ( (0, \pi) ), the function is positive when x is in ( (0, \pi/2) ) and negative when x is in ( (\pi/2, \pi) ), with a zero at ( \pi/2 ). In ( (-\pi, 0) ), it's negative when x is in ( (-\pi/2, 0) ) and positive when x is in ( (-\pi, -\pi/2) ), with a zero at ( -\pi/2 ).

Now let's analyze each graph:

  1. Top-left (blue): Asymptotes at ( -\pi, 0, \pi ). In ( (0, \pi) ): the left part (near 0) is below the x-axis (negative), right part (near ( \pi )) is above (positive). That would mean it's increasing from negative to positive, which is the opposite of ( \cot x ) (which decreases from positive to negative in ( (0, \pi) )). So blue is out.

  2. Top-right (green): Passes through the origin (0,0), but ( \cot 0 ) is undefined (asymptote at x=0), so green is out.

  3. Bottom-left (orange): Passes through the origin (0,0), same issue as green: ( \cot 0 ) is undefined, so orange is out.

  4. Bottom-right (red): Asymptotes at ( -\pi, 0, \pi ). In ( (0, \pi) ): the graph is on the right side of the asymptote at 0 (x=0), going from top (positive) down to bottom (negative) as x approaches ( \pi ), which matches ( \cot x ) (decreasing from ( +\infty ) to ( -\infty ) in ( (0, \pi) ), crossing the x-axis at ( \pi/2 )). In ( (-\pi, 0) ): it's on the left side of the asymptote at 0 (x=0), going from bottom (negative) up to top (positive) as x approaches 0 from the left, which matches ( \cot x ) (increasing from ( -\infty ) to ( +\infty ) in ( (-\pi, 0) ), crossing the x-axis at ( -\pi/2 )). Also, it doesn't pass through the origin, which is correct because ( \cot 0 ) is undefined.

So the bottom-right (red) graph is the correct one. Wait, but let's confirm with another approach: ( \cot x = \frac{1}{\tan x} ), so where ( \tan x ) is positive, ( \cot x ) is positive, and where ( \tan x ) is negative, ( \cot x ) is negative. ( \tan x ) is positive in ( (0, \pi/2) ) and ( (\pi, 3\pi/2) ), negative in ( (\pi/2, \pi) ) and ( (3\pi/2, 2\pi) ) (or ( (-\pi/2, 0) ) and ( (-\pi, -\pi/2) ) for negative x). So ( \cot x ) is positive in ( (0, \pi/2) ) and ( (\pi, 3\pi/2) ), negative in ( (\pi/2, \pi) ) and ( (-\pi/2, 0) ). So in ( (0, \pi) ), ( \cot x ) is positive then negative, which matches the bottom-right graph (positive near 0, negative near ( \pi )). In ( (-\pi, 0) ), ( \cot x ) is negative near 0 (since ( (-\pi/2, 0) ) is negative) and positive near ( -\pi ) (since ( (-\pi, -\pi/2) ) is positive), which also matches the bottom-right graph (negative near 0, positive near ( -\pi )).

So the correct graph is the bottom-right one (red).

Answer:

The graph on the Bottom-Right (the red graph)