# why ezplot(f,[3000,4000]) would not work

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Abdulaziz Abutunis on 28 Feb 2016
Commented: Walter Roberson on 28 Feb 2016
Hi all,
I have this next function which is the result of solving for Px in my code
f=@(Px) (16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27;
the problem is that when I use the ezplot(f,[3000,5000]) it will plot wrong plot. However, if I use
ezplot((16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27,[3000,5000])
the curve will be correct. Please if you have any suggestion to solve this issue advise me.
Thanks
Aziz

John D'Errico on 28 Feb 2016
No. Actually, the result that you claim does work, will fail in MATLAB.
ezplot((16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27,[3000,5000])
Undefined function or variable 'Px'.
I think you have things confused.
Abdulaziz Abutunis on 28 Feb 2016
I am sorry I forgot the ' when I posted the function it should be ezplot('(16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27',[3000,5000])
Thank you, Aziz

Walter Roberson on 28 Feb 2016
Your second form relies upon whatever Px happens to be in memory, and will fail if Px is not a scalar.
Possibly what you meant to post was
ezplot('(16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27',[3000,5000])
When I try that, the output is identical to plotting with your f anonymous function.
When I use your f anonymous function, ezplot gives a warning about f not being vectorized. You can remove that by vectorizing it:
f=@(Px) (16*(-(((27*Px)/2 - 65625).*((27*Px)/2 + 65625))/4).^(1/2))/27;
The output is the same exactly as for the non-vectorized version, and the same exactly as for the string version.