Why matlab gives graph of exp(1/x) and exp(-1/x) wrong?

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ali yaman
ali yaman 2021-5-17
评论： ali yaman 2021-5-17
when trying to find graph of exp(1/x) and exp(-1/x) i am getting graphs that definitely not belong to exp(1/x) and exp(-1/x).
my codes are:
syms x
y=exp(-1/x) % or y=exp(1/x)
fplot(y)
when i run this, i get the graphs that as in below, figure 1 is belong to exp(-1/x) and figure 2 is belong to exp(1/x)
figure 1=exp(-1/x)
figure 2=exp(1/x)
But, this is absolutely ridiculous. The graphs of exp(1/x) and exp(-1/x) must be like :
Is there anybody that will help me? I do not know where i am wrong, please if you know help me
thanks

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Stephen23
Stephen23 2021-5-17
编辑：Stephen23 2021-5-17
Ran in:
fun = @(x) exp(-1./x);
fplot(fun)
ylim([0,10])
fun = @(x) exp(1./x);
fplot(fun)
ylim([0,10])
I doubt there is much that can be done to change the automatic discontinuity detection algorithm, but feel free to read the documentation and see.
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Stephen23
Stephen23 2021-5-17
"Now, how can we decide the limitations, to get a correct shape of curve?"
The shape is always "correct", as explained above. The only difference is how humans like to focus on particular parts or features of the curve: the computer does not know which parts you are interested in, only you know that. Possibly someone has published a heuristic algorithm to predict Y-limits for arbitrary functions.
ali yaman
ali yaman 2021-5-17
Dear @Stephen Cobeldick thanx for all your answer, i got it.
i can't thank you enough!

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