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fert
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The value that makes the derivative zero

Asked by fert
on 17 Jan 2016
Latest activity Edited by John D'Errico
on 17 Jan 2016
Consider the function:
f(x) = x^2* e^(-x)
.
I want to find the point that makes the derivative of this function zero. My code is:
syms x
f=(x.^2).*exp(-x);
k=diff(f);
a = @(x) k;
b=fzero(a,0,1000)
I am getting this error:
"Function value at starting guess must be finite and real."
What do you think it is wrong?

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1 Answer

Answer by John D'Errico
on 17 Jan 2016
Edited by John D'Errico
on 17 Jan 2016
 Accepted Answer

Why not see what MATLAB thinks you are giving it? When you get an error, think about what it is telling you. Then use MATLAB to see what your code does. GET YOUR HANDS DIRTY.
a(0)
ans =
2*x*exp(-x) - x^2*exp(-x)
See that when you created the function handle a, you never told MATLAB that k was actually a function of x. You might have tried this instead:
a = @(x) double(subs(k,x));
a(0)
ans =
0
So now MATLAB can evaluate a as a function of x.
fzero(a,[1,10])
ans =
2
fzero(a,.5)
ans =
7.844e-23
It seems to work now. Of course, simpler was just:
solve(k==0)
ans =
0
2

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