The value that makes the derivative zero

fert (view profile)

on 17 Jan 2016
Latest activity Edited by John D'Errico

John D'Errico (view profile)

on 17 Jan 2016
Accepted Answer by John D'Errico

John D'Errico (view profile)

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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Answer by John D'Errico

John D'Errico (view profile)

on 17 Jan 2016
Edited by John D'Errico

John D'Errico (view profile)

on 17 Jan 2016

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