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Finding full graph / minimum without an x range

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Ashley Sullivan
Ashley Sullivan on 29 Mar 2020
Edited: dpb on 30 Mar 2020
I am plotting a function and looking for its minimum, but the only parameter for the x value that I'm given is that it needs to be in increments of 1. I can also infer from the problem that the x must be positive.
Is there a function to make a "fit" of the graph without knowing an explicit x range? Right now, my method is to just guess for x variables until I get something that looks right, but I'd rather not do it that way.

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dpb
dpb on 29 Mar 2020
No idea what you're asking, sorry...
You can plot versus ordinal number; if the issue is to simply find a minimum in a vector, then plotting isn't even needed.
We need a real example to look at methinks...
Ashley Sullivan
Ashley Sullivan on 29 Mar 2020
To clarify: I need to find the minimum by both plotting and without plotting. Without plotting, I'm just using fminbnd.
Here is my function:
y = 0.04.*x.^2 + 0.3.*(16100./x).^2;
For x, the only information that I am given is that it should be plotted in increments of 1 and that it should be positive (given that it represents velocity). For graphing, I just guessed the interval:
x = 0:1:1000000;
I was wondering if there was an alternative I could put in to prevent me from having to guess.
Mohammad Sami
Mohammad Sami on 30 Mar 2020
y = @(x)0.04.*x.^2 + 0.3.*(16100./x).^2;
fplot(y);
fplot(y,[1 1000]); % to plot a range

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Answers (1)

dpb
dpb on 30 Mar 2020
Edited: dpb on 30 Mar 2020
Well, with such disparate magnitudes in terms, the interesting part is somewhat harder to envision...the second part goes to +inf @ origin while --> 0 as x-->inf so quadratic doesn't have any real counter effect until x gets large enough that the inverse x^2 term begins to go away...
To try to do something automagically to bound, I'd probably try something like
>> fnz=@(x) 0.04*x.^2 - 0.3*(16100./x).^2
fnz =
function_handle with value:
@(x)0.04*x.^2-0.3*(16100./x).^2
>> x0=fsolve(fnz,100)
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
x0 =
209.9802
>> xl=10.^fix(log10(z0)+[-1 1]);
>> yl=10.^fix(log10(z0)+[ 0 1]);
>> fplot(y,xl) % using Mohammed's function definition above here...
>> xlim(xl), ylim(yl)
>> hAx=gca;
>> hAx.XScale='log';
>> hAx.YScale='log';
>> grid on
produced

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