Graph question (1 Viewer)

[Saturn WY15]

Can some explain the graph y=x/arctanx

 

[4025808]

Can some explain the graph y=x/arctanx

it is an even function, that's the first part.

Hint: Draw arctanx, 1/arctanx and x on the same axes. Then multiply the ordinates of 1/arctanx and x together.

 

[Saturn WY15]

How do u even draw 1/arctanx

 

[Shadowdude]

How do u even draw 1/arctanx

You use the graph for arctan(x)

And then take reciprocals of ordinates.

 

[Saturn WY15]

How about when x=0 therefore arctan=0, what happens to the reciprocal.

 

[Shadowdude]

How about when x=0 therefore arctan=0, what happens to the reciprocal.

It goes off to infinity...

 

[Saturn WY15]

0/infinity= 0 that's wrong, cause when x=0 on graph x/arctanx y=1

 

[RealiseNothing]

0/infinity= 0 that's wrong, cause when x=0 on graph x/arctanx y=1

Use L'Hopital's rule:

 

[Aesytic]

it's also an open circle when x=0

 

[RealiseNothing]

it's also an open circle when x=0

No it's not, when x=0, f(x)=1.

 

[RealiseNothing]

0/infinity= 0 that's wrong, cause when x=0 on graph x/arctanx y=1

The reciprocal of arctanx when x=0 does go off to infinity. The reason why x/arctanx when x=0 is 1 is due to the 'x' that's now in the numerator, which changes the graph.

 

[lolcakes52]

Look, at the point x=0 y=arctan(x) has a gradient of 1 and y=x has a gradient of 1 and they both have a value of y=0. SO AT THE POINT x=0 THEY ARE BASICALLY THE SAME GRAPH! So they are equal at that point and the ratio between the two is 1, comprendai?

 

[Aesytic]

when x=0, x/arctanx = 0/0 which is undefined
i know the limit as x approaches 0 is 1, but because it is undefined when x=0, shouldn't it be an open circle at (0,1)?

 

[RealiseNothing]

when x=0, x/arctanx = 0/0 which is undefined
i know the limit as x approaches 0 is 1, but because it is undefined when x=0, shouldn't it be an open circle at (0,1)?

If the numerator wasn't 0 it would be, but because it is a 0 it changes the graph. ie 1/0, 2/0 etc would have an open circle, but 0/0 doesn't necessarily.

 

[Sy123]

I graphed f(x)=x/atan(x) on Geogebra. Then I put in f(0), the result said undefined, so therefore there should be an open circle there?

 

[seanieg89]

There should be an open circle at x=0. The expression 0/0 is undefined, so the function we are sketching is not defined at x=0. That the limit of f(x) is 1 as x approaches 0 is irrelevant.

Eg, the graph of y=x/x is the constant line y=1 with an open circle at the point (0,1).

Look up the phrase "removable singularity" if you want more examples.

 

[RealiseNothing]

Hmm I graphed it as well and it didn't come up with an open circle, oh well.

 

[seanieg89]

Some graphing software will 'simplify' the expression before graphing it, this can cause some domain problems.

 

[Carrotsticks]

Existence of left/right limit doesn't necessarily guarantee continuity.

 

[RealiseNothing]

Existence of left/right limit doesn't necessarily guarantee continuity.

Yer, I was thinking that the fact the numerator also went to 0 had some impact, guess not lol.