Thread: Computing the value of a function

Hi there,

I'm having trouble getting the desired output. whenever I enter a number, my program just freezes and doesn't do anything. if it does do something, it's always the number I input. e.g: if I put in 2, it will give me 2 in return.

Here's the code

Code:

#include <stdio.h>
#include <math.h>


int main(void)
{
  double x;


  printf("Enter a number x: ");
  scanf("%lf\n", &x);
  if ((x*x-5*x) == 0)  {
    printf("f(x) is not defined\n");
    }
  else if ((3*x + 6) < 0)  {
    printf("f(x) is not defined\n");
    }
  else (((1+(x*x)/sqrt(3*x+6))/(x*x-5*x)));  {
    printf("f(x) = %lf\n", x);
    }
  return(0);
}

Should I implement a function to call on? any other type of loop maybe?

Please help.

Ok, so I revamped my code a little bit:

Code:

#include <stdio.h>
#include <math.h>


double function(double x);


int main()
{
  double x;


  printf("Enter a number x: ");
  scanf("%lf\n", &x);


  if (x != 0 || x > -2)  {
    printf("f(x) = %lf\n", function(x));
    }
  else   {
    printf("f(x) is not defined\n");
    }
  return(0);
}

double function(double x)
{
 return (((1+(x*x)/sqrt(3*x+6))/(x*x-5*x)));
}

So here's my problem now: It will ask me to enter a value of x, and I do. but I have to enter another number after it in order for the program to give me output. This is what it looks like:

Code:

Enter a number x: 2.74 //This doesn't do anything
2.74 //This is the number that gets the program to give output
f(x) = -0.482997

What am I doing wrong?