%%
%Cerinta 3.a
f=@(x) exp(x)-2;
X=linspace(-2,3,100);
Y=f(X);
plot(X,Y,'linewidth',2);

hold on

g=@(x) cos(exp(x)-2);
XX=linspace(-2,3,100);
YY=g(XX);

plot(XX,YY,'linewidth',2);
xL=xlim;
yL=ylim;
line(xL,[0 0],'Color','k','Linewidth',0.01) %axa ox
line([0 0],yL,'Color','k','Linewidth',0.01) %axa oy
%%
%Cerinta 3.b
%Pentru a afla solutia ecuatiei, vom afla de fapt solutia ecuatiei:
%e^x -2 -cos(e^x -2)  h(x)=f(x)-g(x)

h=@(x) exp(x)-2 -cos(exp(x)-2);
eps=10^(-5);
[solutie]=MetBisectie(h,0.5,1.5,eps)
%%
%Cerinta 8.a
f=@(x) x.^3 - 18*x -10;
X=linspace(-5,5,100);
Y=f(X);
plot(X,Y,'linewidth',2);

grid on
hold on
xL=xlim;
yL=ylim;
line(xL,[0 0],'Color','k','Linewidth',0.01) %axa ox
line([0 0],yL,'Color','k','Linewidth',0.01) %axa oy

%Cerinta 8.d
%Vom alege urmatoarele intervale: [-5,-3];[-2,0];[3,5]
%Functia este monotona pe aceste intervale, ceea ce inseamna ca derivata
%e diferita de 0. 
%Avem ca f(-5)<0 si f(-3)>0 => f(-5)*f(-3)<0
%Avem ca f(-2)>0 si f(0)<0 => f(-2)*f(0)<0
%Avem ca f(3)<0 si f(5)>0 => f(3)*f(5)<0


f=@(x) x.^3 - 18*x -10;
eps=10^(-3)
%Aflam cele 3 aproximari ale solutiilor
[xaprox1]=MetSecantei(f,-5,-3,-4.9,-3.1,eps);
[xaprox2]=MetSecantei(f,-2,0,-1.9,-0.1,eps);
[xaprox3]=MetSecantei(f,3,5,3.1,4.9,eps);

plot(xaprox1,f(xaprox1),'o','MarkerFaceColor','g','MarkerSize',10)
text(-3.6,-4,'{xaprox1}','EdgeColor','y','LineWidth',1.5)
plot(xaprox2,f(xaprox2),'o','MarkerFaceColor','g','MarkerSize',10)
text(-2,-4,'{xaprox2}','EdgeColor','y','LineWidth',1.5)
plot(xaprox3,f(xaprox3),'o','MarkerFaceColor','g','MarkerSize',10)
text(3.1,5,'{xaprox3}','EdgeColor','y','LineWidth',1.5)
xlabel('x');    
ylabel('y');
hold off

%%
%Cerinta 8.e
%Alegem intervalele: 
%[-4.5,-3.5], avem o singura solutie (graficul functiei intersecteaza OX intr-un singur punct)
%[-1,5,1]
%[4,5]

f=@(x) x.^3 - 18*x -10;
X=linspace(-5,5,100);
Y=f(X);
plot(X,Y,'linewidth',2);

grid on
hold on
xL=xlim;
yL=ylim;
line(xL,[0 0],'Color','k','Linewidth',0.01) %axa ox
line([0 0],yL,'Color','k','Linewidth',0.01) %axa oy


eps=10^(-3);
%Aflam cele 3 aproximari ale solutiilor
[xaprox1]=MetPozFalse(f,-4.5,-3.5,eps);
[xaprox2]=MetPozFalse(f,-1.5,1,eps);
[xaprox3]=MetPozFalse(f,4,5,eps);

plot(xaprox1,f(xaprox1),'o','MarkerFaceColor','g','MarkerSize',10)
text(-3.6,-4,'{xaprox1}','EdgeColor','y','LineWidth',1.5)
plot(xaprox2,f(xaprox2),'o','MarkerFaceColor','g','MarkerSize',10)
text(-2,-4,'{xaprox2}','EdgeColor','y','LineWidth',1.5)
plot(xaprox3,f(xaprox3),'o','MarkerFaceColor','g','MarkerSize',10)
text(3.1,5,'{xaprox3}','EdgeColor','y','LineWidth',1.5)
xlabel('x');    
ylabel('y');
hold off


%%
%Exercitiul 7.a
%Avem ecuatia x^3+4x-1=0 x in [0,1]
%f(x)=x^3+4x-1
%Avem derivata functiei f'(x)=3x^2 +4
%f'(x) este strict pozitiva pe intervalul [0,1]  (1)
%f'(x)=0 nu are solutie reala pe intervalul [0,1] (2)
%din (1),(2) rezulta ca f(x) este strict crescatoare pe intervalul [0,1]
%daca f(x) e strict crescatoare pe [0,1], atunci graficul lui f taie
%axa Ox intr-un singur punct, rezulta ca f(x)=0 are solutie unia pe [0,1]



%%