enter image description hereSo I have this function :
$f(x)= e^{((x²+1)/(x-1))}$
I want to draw the function using what's possible.
1) the domain is $(-∞,1) \cup (1,∞)$
2) the $x$ intercept doesn't exist, whilst the $y$ intercept is $1/e$.
3) The asymptotes: vertical asymptote is where $x=1$ and horizontal asymptote is where $y=0$
4) I compute the derivative: I find the critical point is $x=1$, and the $x$ 's where the function is equal to $0$ , which is $x=1+\sqrt{2}$ and $x=1-\sqrt{2}$. As the critical point $x=1$ isn't part of the domain , should I use it to find min/max of the function ?
With the intervals, I find out that the function has a local min at $x=1+\sqrt{2}$ and a local max at $x=1-\sqrt{2}$.
If I use all this information to draw a graph , I get something kind of weird, especially due to the vertical asymtpote which I found out was equal to $1$.
I am not sure how to draw the graph , so I plot it into my ti nspire cx cas calculator and the image I attached shows the function . However, I don't see a vertical asymptote , but what I do see is a sort of $x$ interecept on the $x$ axis , at about $1$. I don't even see a minimum , to be honest. How come ? I am confused about the minimum and , especcialy , the asymtpotes. Is there somethig wrong with my calculations? Is the graph depicted on my calculator correct?
If i plot the same function on symbolab, i get a completely different equation... how come?
Close to $x=1$ the function is almost infinity; so you cannot see it.
Plor again for $1.5 \leq x \leq 3.5$ and you will see the minimum.