I'm supposed to find all extremum to the function f(x)=((1+x​^2−1.6x^3+​0.6x^4)/(1​+x^4)). Been struggeling with this for hours.

2 vues (au cours des 30 derniers jours)
I'm supposed to find all extremum to the function f(x)=((1+x^2−1.6x^3+0.6x^4)/(1+x^4)). Been struggeling with this for hours.
  1 commentaire
James Tursa
James Tursa le 7 Nov 2016
What have you tried so far? Is your code having errors, not producing correct answers, or what?

Connectez-vous pour commenter.

Réponses (1)

Hannes Daepp
Hannes Daepp le 11 Nov 2016
I understand that you want to find the minima and maxima for this function. From a plot, you can observe that this function has two global extremum, limits at x=+/-inf, and two local max/min:
>> x=-10:0.1:10;
>> f=((1+x.^2-1.6*x.^3+0.6*x.^4)./(1+x.^4));
>> plot(x,f)
You could determine absolute min/max numerically:
>> [ma, imax] = max(fvec);
>> [mi, imin] = min(fvec);
>> disp([xvec(imax) ma])
-1.1000 2.1176
>> disp([xvec(imin) mi])
1.9000 0.1037
Alternatively, you can use the Symbolic Toolbox to find limits and extremum: >> syms x >> f=((1+x^2-1.6*x^3+0.6*x^4)/(1+x^4)); >> limit(f,x,-inf) ans = 3/5
Minima and maxima can be found by setting the derivative equal to zero. In this case, the output has complex roots, so some extra steps must be undertaken to determine the real roots:
>> syms x f(x)
>> f=((1+x^2-1.6*x^3+0.6*x^4)/(1+x^4));
>> limit(f,inf)
ans =
3/5
>> rts=solve(diff(f,x))
rts =
0
root(z^5 - (5*z^4)/4 - z^2 - 3*z + 5/4, z, 1)
root(z^5 - (5*z^4)/4 - z^2 - 3*z + 5/4, z, 2)
root(z^5 - (5*z^4)/4 - z^2 - 3*z + 5/4, z, 3)
root(z^5 - (5*z^4)/4 - z^2 - 3*z + 5/4, z, 4)
root(z^5 - (5*z^4)/4 - z^2 - 3*z + 5/4, z, 5)
This function has several roots beyond those at zero. To find those locations, we can use "roots" and pick out the real roots:
>> roots([1 -5/4 0 -1 -3 5/4])
ans =
1.8824 + 0.0000i
-1.0923 + 0.0000i
0.0466 + 1.2870i
0.0466 - 1.2870i
0.3666 + 0.0000i

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by