Make a tunction that calculates the CUBE-ROOT of the input number and returns the value (single or multiple). Cube root of 125 = 5.00000 Roots can be both real or complex #include You can not use math.h You can not use loops (for do while) double cubeRoot(double n) { double i, precision = 0.000001; for(i = 1; (i*i*i) <= n; ++i); //Integer part for(--i; (i*i*i) < n; i += precision); //Fractional pa You can modify this code: return i; Hint: int main() { int n = 125; N 1 2x, + 3 Xn+1 = x. printf("Cube root of %d = %lf", n, cubeRoot(n)); z

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Make a function that calculates the CUBE-ROOT of the input
number and returns the value (single or multiple).
Cube root of 125 = 5.000000
Roots can be both real or complex
#include <stdio.h>
You can not use math.h
double cubeRoot(double n) {
double i, precision = 0.000001;
You can not use loops (for do while)
for(i = 1; (i*i*i) <= n; ++i);
//Integer part
for(--i; (i*i*i) < n; i += precision); //Fractional part
You can modify this code:
return i;
}
int main() {
Hint:
int n = 125;
N
1
2x, +
Xn+1
printf("Cube root of %d = %lf", n, cubeRoot(n));
return 0;
Transcribed Image Text:Make a function that calculates the CUBE-ROOT of the input number and returns the value (single or multiple). Cube root of 125 = 5.000000 Roots can be both real or complex #include <stdio.h> You can not use math.h double cubeRoot(double n) { double i, precision = 0.000001; You can not use loops (for do while) for(i = 1; (i*i*i) <= n; ++i); //Integer part for(--i; (i*i*i) < n; i += precision); //Fractional part You can modify this code: return i; } int main() { Hint: int n = 125; N 1 2x, + Xn+1 printf("Cube root of %d = %lf", n, cubeRoot(n)); return 0;
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