1) de la función hallar su grafica y el cuadro de la bisseccion

b.

function [c,iter] = bissec(a,b,e)iter=0; fprintf(' ==========================================================\n') fprintf(' iter a b c f(a) f(c) \n') fprintf(' ==========================================================\n')while abs(b-a) > e c=(a+b)/2; fprintf('%5d %10.6f %10.6f %10.6f %10.6f %10.6f \n', iter,a,b,c,f(a),f(c)); if f(a)*f(c) > 0 a=c; else b=c; end iter=iter+1;end; end

function y = f(x)y=(x*exp(2))-1;end

>> [c,iter] = bissec(1,2,0.001)

==========================================================

iter a b c f(a) f(c)

==========================================================

0 1.000000 2.000000 1.500000 6.389056 10.083584

1 1.500000 2.000000 1.750000 10.083584 11.930848

2 1.750000 2.000000 1.875000 11.930848 12.854480

3 1.875000 2.000000 1.937500 12.854480 13.316296

4 1.937500 2.000000 1.968750 13.316296 13.547204

5 1.968750 2.000000 1.984375 13.547204 13.662658

6 1.984375 2.000000 1.992188 13.662658 13.720385

7 1.992188 2.000000 1.996094 13.720385 13.749249

8 1.996094 2.000000 1.998047 13.749249 13.763680

9 1.998047 2.000000 1.999023 13.763680 13.770896

c =1.9990

iter = 10

2) demostrar la formula de la escalera

function [x, iter] = newtonraphson (x, e)iter = 0;fprintf ( ' =============\n')fprintf ( ' iter x \n')fprintf ( ' =============\n')while abs (f(x)) > efprintf ( ' %5d %10.6f \n', iter, x);x = x - f(x)/df(x);iter = iter + 1;if iter >1000error ( 'parece que no converge newton' );endendend

function y = f(x)y=x^4-20*x^3+700*x^2-14000*x+70000;

function y=df(x) y=4*x^3-60*x^2+1400*x-14000;end

>> [x, iter] = newtonraphson (12, 0.000001) ============= iter x ============= 0 12.000000 1 22.283582 2 18.149899 3 15.872773 4 15.033612 5 14.911875 6 14.909370

x =14.9094iter = 7

3) hallar la viga y hallar las iteraciones: por el método de la bisección.

function y = f(x)y=-185*x+1650;

>> [c,iter] = bissec(8,10,0.000001)

c =8.9189iter = 21