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?

  • $\begingroup$ Yes it is . E to the power of (x²+1)/(x-1) $\endgroup$ Jun 17, 2018 at 13:14

1 Answer 1


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.

  • $\begingroup$ why did you choose these values ? $\endgroup$ Jun 17, 2018 at 13:23
  • $\begingroup$ @MichaelMorello. Just because $1+\sqrt 2 \approx 2.5$. Then $\pm 1$ to get an inverval. $\endgroup$ Jun 17, 2018 at 13:25
  • $\begingroup$ If i choose the domain to be the values you said, i wouldn't see anything.. because half of the graph belongs to the negative part of the axis. maybe -1.5? $\endgroup$ Jun 17, 2018 at 13:25
  • $\begingroup$ What about the aysmptotes? can't see any vertical one.... $\endgroup$ Jun 17, 2018 at 13:27
  • $\begingroup$ Moreover, if i plot the same function on symbolab, i get a completely different graph... $\endgroup$ Jun 17, 2018 at 13:28

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