undefined at pi/2 ?

[anders]

Is the function f(x) = (cosx)(secx) defined at x=pi/2 ?

 

[Deleted member 4993]

Is the function f(x) = (cosx)(secx) defined at x=pi/2 ?

No .... plot it and observe!....... edited

The function f(x) reaches the same limit from left and from the right. Since sec(x) is DNE at x = pi/2, the function does not exist there!

Try plotting (x^2 -4)/(x -2). You know there is a "removable" discontinuity at x = 2. However, plotting softwares will show a continuous curve through x = 2

 

[anders]

My plotting software may be incompetent (or it may be user error!), but all I see is y=1 with no discontinuities.

 

[pka]

Is the function f(x) = (cosx)(secx) defined at x=pi/2 ?

Yes it is not defined at \(\displaystyle x=\frac{\pi}{2}\) SEE HERE

 

[anders]

I believe you, but no plotting software that I have used so far is wiling to show a 'donut' at pi/2. They show f(pi/2)=1. Not very satisfying. Is there any plotting routine out there that can show me the truth? (I can handle it! )

 

[anders]

No .... plot it and observe!....... edited

The function f(x) reaches the same limit from left and from the right. Since sec(x) is DNE at x = pi/2, the function does not exist there!

So, the limit exists (equals 1), but the function DNE. I believe that. I just can't find any plotting software that admits it!

 

[Deleted member 4993]

My plotting software may be incompetent (or it may be user error!), but all I see is y=1 with no discontinuities.

My response in #2 was incorrect (incomplete). I fixed it.

Your plotting software is quite competent - whereas I cannot say the same about my fingeres!

No .... plot it and observe!....... edited

 

[anders]

No .... plot it and observe!....... edited

The function f(x) reaches the same limit from left and from the right. Since sec(x) is DNE at x = pi/2, the function does not exist there!

Try plotting (x^2 -4)/(x -2). You know there is a "removable" discontinuity at x = 2. However, plotting softwares will show a continuous curve through x = 2

Thanks for the removable discontinuity example! So I'm not going crazy (or, at least not for this reason!).

 

[Dr.Peterson]

I have wished for graphing software that would show removable discontinuities, but haven't found one yet. (To plot such things in teaching, I have to manually insert the open dot, and students need to be told to do so themselves.) If anyone does know of something like that, I'd like to know!

Desmos does show an open dot when I click on its location, but not otherwise.

And I found this page explaining some of the issues in the case of Maple. If you think about it, graphers typically just plot points, dumbly; to identify discontinuities and give them special treatment requires intelligence.

And with that in mind, I discovered that Wolfram Alpha can do it, when I ask explicitly.

 

[anders]

I have wished for graphing software that would show removable discontinuities, but haven't found one yet. (To plot such things in teaching, I have to manually insert the open dot, and students need to be told to do so themselves.) If anyone does know of something like that, I'd like to know!

Desmos does show an open dot when I click on its location, but not otherwise.

And I found this page explaining some of the issues in the case of Maple. If you think about it, graphers typically just plot points, dumbly; to identify discontinuities and give them special treatment requires intelligence.

And with that in mind, I discovered that Wolfram Alpha can do it, when I ask explicitly.

Thanks Dr. Peterson. I would like my students to show a couple donut holes in the interval 0 <= x <= 2pi .

 

[Dr Ahkim]

Is the function f(x) = (cosx)(secx) defined at x=pi/2 ?

The question is whether [MATH]f(x)=\cos x \cdot \sec x[/MATH] is defined at [MATH]x=\frac{\pi}{2}[/MATH]. Since [MATH]f(x)[/MATH] is a product of two functions, we need to look at the factors. There is no problem with [MATH]\cos \frac{\pi}{2}=0[/MATH], however [MATH]\sec(x)=\frac{1}{\cos x}[/MATH], it follows that [MATH]\sec \frac{\pi}{2}[/MATH] is not defined.

There is no software which will plot a donut hole, because such a hole takes room and would imply the function is not defined in an interval around [MATH]\frac{\pi}{2}[/MATH]. This is not true. I remember also stumbling on this before

 

[Harry_the_cat]

I know that Casio graphics calculators will show point discontinuities if you zero in using a reduced viewing window. Haven't tried y = cos(x).sec(x) myself but I have for graphs such as y = (x^2-1)/(x-1) and there is a visible break in the line.