@robjohn hello sir, I am a big admirer of your answers

sir, if it's okay with you, can you please explain how the intermediate value theorem justifies your argument?

i.e tanx=x^2+1 guarantees continuity at that point?

what post are you looking at ?

oh, I see

If you look at the function $\tan(x)-x^2-1$ at the ends of each if those intervals, it is negative at the left and positive at the right

So it has to be $0$ somewhere in between

@satan29: we are using that the function is continuous on those intervals.

I am off to walk that dogs. BBL

o/

@robjohn i get that, but how does that relate to continuity of the function in question

oh I see, you were using the IMVT to just justify that tan x = x^2+1 at 4 points

i thought you used the IMVT to justify that that tanx=x^2+1 implies that the original(i.e the piecewise one) function is continuos

Yes. I had to find the original question to see what this was all about.

we are looking for where $x^2+1$ for $x\in\mathbb{Q}$ and $\tan(x)$ for $x\not\in\mathbb{Q}$ is continuous

yes

Sorry, it took me a while to find the PDF that had the question

no worries sir, I am grateful that you are taking out the time to address this

@robjohn yes.

i really struggle with the idea of "approach" when functions are defined like this @robjohn

@satan29 Yeah. In this case, we just care about when the two functions coincide

but sir how does that guarantee continuity?

since both $x^2+1$ and $\tan(x)$ are continuous, at those points where $x^2+1=\tan(x)$, the limit of $f$ exists.

and the value of $f$ equals that limit

a function is continuos at x=x0 if lim(h-->0+)f(x0+h)=lim(h-->0-)f(x0+h)=f(x0)

yes

sir can we prove your conjecture using this definition?

yes. since $x^2+1$ is continuous and $\tan(x)$ are continuous, the limit statement holds for them.

ohhh hmmm

Thus, at those points where $x^2+1=\tan(x)$, the limits are equal and thus, $f$ also satisfies the limit requirements for continuity

whether your points are in $\mathbb{Q}$ or not

@robjohn sir can i plot this function in wolfram alpha?

I am not sure, since it is different on $\mathbb{Q}$ and off

yeah..

You can plot the two functions and know that on $\mathbb{Q}$ it is one and off it is the other

any software that can help me plot this that you know of?

@satan29 the problem is that $\mathbb{Q}$ is dense in $\mathbb{R}$, so no plotting software will be able to separate the points.

hmm

That is about as close as you'll get

yellow when $x\in\mathbb{Q}$, blue otherwise

but no matter, as the function approaches any of the points where the continuous curves cross, the function that flips between the two, depending on $x\in\mathbb{Q}$, is continuous

Because both continuous curves tend to the same value at those points.

yes, that makes sense

thank you so much @robjohn

you're welcome

It's good to see someone trying to understand rather than just trying to get answers.

:-)

it's really satisfying when it all kinda clicks

o/