how to find max vaule of function handle?

f=@(u) (u^2/(1+u^2))*(2/(u^2+0.55))-u

[Walter Roberson]

The maximum is an irrational number involving one of the roots of a degree 8 polynomial.

Answer 1

[John D'Errico]

help fminbnd
help fminsearch

Now, you might be confused, as those are minimization tools. But suppose you multiplied your function by -1? Would the location now of a minimum be in the same location as the maximum you really want?

Do you know how to create a new function that is the same as another, but just negated? What if you did this?

fneg = @(u) -f(u);

[John D'Errico]

Note that this function only has a local maximum. In fact, a global maximum does not exist.

syms u
F = (u^2/(1+u^2))*(2/(u^2+0.55))-u;
ezplot(F,[-5,5])

You can easily show the function itself is asymptotically linear as u approaches +inf or -inf. In order to find that local max, you will need to be careful in your choice of starting values. Or, you can use the symbolic toolbox, though you cannot find an analytical form for that local max.

[Walter Roberson]

The analytical location of the local max is

root(z^8 + (31*z^6)/10 + 4*z^5 + (1401*z^4)/400 + (341*z^2)/200 - (11*z)/5 + 121/400, z, 2)

but there is no algebraic formula for that