Where does the graph of arctan x have horizontal asymptotes?

Question

I have an assignment with these choices:

0
-pi/2
nowhere
pi/2

I think it is 'nowhere'....is this correct? Thanks.

Answer 1

tan x has VERTICAL asymptotes at ±π/2, so that's where its inverse will have HORIZONTAL asymptotes.

Make a backwards L with your right thumb pointing left and your right index finger pointing up. Make an upside down L with your left thumb pointing right and your left index finger pointing down. Touch thumbs. Your thumbs and fingers make a somewhat squared off graph of tan x, thumbs touching at origin. Now keeping thumbs touching, reverse direction the fingers point and rotate the whole thing 1/4 turn right so the left index finger points right. Now you have a somewhat squared off graph of arctan x. Depending on scale, arctan is a lazy S.

Answer 2

Graph Of Arctan

Answer 3

at π/2 and -π/2

y = arctan(x) can also be expressed as y = tan‾¹x or x = tan(y)

so all you are doing is interchanging the x and y values of the graph y = tan x.

y = tan x has vertical asymptotes at x = π/2 and -π/2 so
x = tan y has horizontal asymptotes at y = π/2 and -π/2

Answer 4

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RE:
Where does the graph of arctan x have horizontal asymptotes?
I have an assignment with these choices:

0
-pi/2
nowhere
pi/2

I think it is 'nowhere'....is this correct? Thanks.