Question

Write a program to print graph of a function on text terminal. Note that the function is given in the code. Therefore you are supposed to call a function whose prototype is double func (double in). A menu appears initially to select one of the five functions given below. User is expected to enter the range and step next. User will keep on entering a new function selection until Exit is selected. Note that depending on the function or/and selected range and step, mimimuim and maximum values to print would be different. Since you have a limited number of columns (80 columns) to print the range of the graphics should be adjusted dynamically with respect to the selected function given range. Also note that you may not produce zero for a given range and step. In such a case if | x- step 1/2 < step, it can be accepted as zero to print y axis. Note that your name and id should be included in the source code file. Sufficent inline coding is also expected.

I will upvote if u will solve
Thank you for your time

Write this code in c

Answer 1

#include<stdio.h>
#include<math.h>
double func1(double x)
{
return x;
}
double func2(double x)
{
return x*(-1);
}
double func3(double x)
{
return x*x;
}
double func4(double x)
{
return x*x*x*(-1);
}
double func5(double x)
{
return (3*cos(x)-(2*sin(4*x))-1)/2;
}
int main()
{
double from,to,interval,min=0,max=0,minx,maxx,result,a;
int choice;
do
{
printf("1.f(x)=x\n2.f(x)=-x\n3.f(x)=x^2\n4.f(x)=-x^3\n5.f(x)=(3*cos(x)-2*(4*x)-1)/2\n0.Exit\n");
printf("Selection: ");
scanf("%d",&choice);
if(choice==0)
return 0;
printf("From: ");
scanf("%lf",&from);
printf("To: ");
scanf("%lf",&to);
printf("Step: ");
scanf("%lf",&interval);
switch(choice)
{
case 1:a=from;
while(a<=to)
{
result=func1(a);
if(result<=min)
{
min=result;
minx=a;
}
if(result>=max)
{max=result;
maxx=a;
}
a+=interval;
}
printf("Min y=%lf for x=%lf\n",min,minx);
printf("Max y=%lf for x=%lf",max,maxx);
a=from;
printf("\n\tx\tf(x)\n");
while(a<=to)
{ result=func1(a);
printf("%lf\t%lf\n",a,result);
a+=interval;
}
break;
case 2:a=from;
while(a<=to)
{
result=func2(a);
if(result<=min)
{
min=result;
minx=a;
}
if(result>=max)
{max=result;
maxx=a;
}
a+=interval;
}
printf("Min y=%lf for x=%lf\n",min,minx);
printf("Max y=%lf for x=%lf",max,maxx);
a=from;
printf("\nx\tf(x)\n");
while(a<=to)
{ result=func2(a);
printf("%lf\t%lf\n",result,a);
a+=interval;
}
break;
case 3:a=from;
while(a<=to)
{
result=func3(a);
if(result<=min)
{
min=result;
minx=a;
}
if(result>=max)
{max=result;
maxx=a;
}
a+=interval;
}
printf("Min y=%lf for x=%lf\n",min,minx);
printf("Max y=%lf for x=%lf",max,maxx);
a=from;
printf("\nx\tf(x)\n");
while(a<=to)
{ result=func3(a);
printf("%lf\t%lf\n",result,a);
a+=interval;
}
break;
case 4:a=from;
while(a<=to)
{
result=func4(a);
if(result<=min)
{
min=result;
minx=a;
}
if(result>=max)
{max=result;
maxx=a;
}
a+=interval;
}
printf("Min y=%lf for x=%lf\n",min,minx);
printf("Max y=%lf for x=%lf",max,maxx);
a=from;
printf("\nx\tf(x)\n");
while(a<=to)
{ result=func4(a);
printf("%lf\t%lf\n",result,a);
a+=interval;
}
break;
case 5:a=from;
while(a<=to)
{
result=func5(a);
if(result<=min)
{
min=result;
minx=a;
}
if(result>=max)
{max=result;
maxx=a;
}
a+=interval;
}
printf("Min y=%lf for x=%lf\n",min,minx);
printf("Max y=%lf for x=%lf",max,maxx);
a=from;
printf("\nx\tf(x)\n");
while(a<=to)
{ result=func5(a);
printf("%lf\t%lf\n",result,a);
a+=interval;
}
break;
default:printf("Invalid Input");
}
}while(1);
return 0;
}

This is the program for the above question but the sample output provided by you does not match with the expression for the fifth case i.e (3*cos(x)-2*sin(4*x)-1)/2.I even tried manually and the answers matched with the answer generated by the program but are different from those provided in the sample output.Some brackets may be missing in the expression,just change the expression in the func5 accordingly for the right answer.Other than that everything else is working fine.