Have your program find which of the y values is closest to 10 (either larger or smaller).
have the program print the x value that gives this closest y value. Also, print how close
the y value is to 10.
int main ()
float x, y;
cout<<"\t\t\tOutput of First Program!"<<endl<<endl;
cout<<"\t\t\t\tValue of X:\t\tValue of Y:"<<endl;
for (x=-3; x<=4; x+=0.5)
y=(9*x*x*x-27*x*x-4*x+12)/(sqrt(3*x*x+1) + abs(5-x*x*x*x));
cout<<"Y is Zero ";
cout<<"Y is Positive ";
cout<<"Y is Negative ";
cout<<"\t\t\tX= "<<x<<"\t\t\tY= " <<y<<endl;
cout<<endl<<endl<<"The program is halting";
How do I go about doing these? Could you please give me a hint but not give me the answer. Thanks for help.
So, you have an equation 10 = f(x) for which you don't have analytic solution.
You have to make a guess; pick some value of x. Calculate f(x). If it is not spot on, then pick a new x and repeat. The question is, how to pick the x wisely so that you do get closer to the correct answer? There are plenty of algorithms for that.