If n were negative, the driver can get the correct answer via function p by using the line of code:

Question

```
1 float p( float x, int n )
2 {
3 if ( n == 0 )
4 return 1;
5 else
6 return x p( x, n – 1 );
7 }
```
A. return 1 / p( x, -n );
B. return 1 / p( x, n );
C. return p( x, -n );
D. return p( x, n );

Answer 1

6