Graphs (1 Viewer)

[omgmynameislong]

Say you were given a graph like cosecx and they tell you to draw 1/f(x), will there be open circles the intercepts or not?

eg. They give you some parabola looking thing like cosecx called f(x) with asymptotes, will there be open circles for intercepts for 1/f(x) or none like sinx?

 

[Trebla]

There will be open circles as the graph is undefined at those points.

 

[omgmynameislong]

It's not really undefined :|, its 1/infinite which should be zero but technically you can't reach it. I dunno.

1/(cscx) - Wolfram|Alpha

1/cosecx gives sinx without any opened dots so ?_?...

 

[lychnobity]

It's not really undefined :|, its 1/infinite which should be zero but technically you can't reach it. I dunno.

1/(cscx) - Wolfram|Alpha

1/cosecx gives sinx without any opened dots so ?_?...

Listen to Trebla.

1/infinity does not equal zero. It comes CLOSE TO IT; but never reaches it. That's why the 'intercepts' should have open circles, as it is defined at every point before and after it, except for the actual intercept.

 

[Rezen]

But in this case, 1/f(x) is cosx which is defined for for all x. If however you where to do f(x)=lnx, then there would be an open circle at x=0.

Atleast this is my understanding of what your ment to graph?

 

[Trebla]

It's not really undefined :|, its 1/infinite which should be zero but technically you can't reach it. I dunno.

1/(cscx) - Wolfram|Alpha

1/cosecx gives sinx without any opened dots so ?_?...

This is where one needs some strict technical precision:
cosec x = 1/sin x ONLY wherever sin x is non-zero
So technically one should write

DEFINITION:
cosec x = 1/sin x (with restriction sin x =/= 0)
By rearrangement
sin x = 1/cosec x (with restriction sin x =/= 0)
The restriction doesn't go away

What I'm basically trying to say is that sin x = 1/cosec x only if sin x is non-zero because of the restriction applied on cosec x which isn't automatically removed just for the sake of it.

You can only do this:
1/cosec x = 1/(1/sin x) = sin x
when sin x is non-zero due to the denominator in the definition of cosec x

 

[cutemouse]

Well by definition, the cosec function has gaps in its domain (namely when sinx=0).

So if f(x)=cosecx then sketching y=1/f(x) will mean that there would indeed be open circles at the x-intercepts as they are not included in the domain.

Also talking about whether the cosec function is defined at the points where sinx=0 doesn't make any sense because they are not included in the domain of the cosec function.