The program below contains an incomplete recursive function raised$\_$to$\_$power$().$

The function returns the value of the first parameter numberof type floatraised to

the value of the second parameter power of type int for all values of power greater

than or equal to $0.$

The algorithm used in this question to write a recursive function to raise a float

value number to a positive power uses repeated multiplication as follows:

numberpower $=1$ if power$=0$

$=$ number x numberpower$-1$ otherwise

In other words, number raised to power gives $1$ if power is $0$;

and otherwise numberpower can be calculated with the formula:

number x numberpower$-1$

$1.$ #include using namespace std;

$2.$ float raised$\_$to$\_$power$()$

$3.\{$

$4.$ if $($power $<0)$

$5.\{$

$6.$ cout $<<"$

Error $-$ can't raise to a negative power

$"$;

$7.$ exit$(1)$;

$9.\}$

$10.$ else if $()$

$11.$ return $()$;

$12.$ else

$13.$ return $($number $*$ raised$\_$to$\_$power$($number$,$ power $-1))$;

$14.\}$

$15.$ main$()$

$16.$

$17.$ float answer $=$ raised$\_$to$\_$power$(4\mathrm{.}0,3)$;

$18.$ cout $<<$ answer;

$19.$ return $0$;

$20.\}$

$($b$)$ Using the fact that any value raised to the power of $0$ is $1,$ complete the base case in line $10$ and $11.$

This question is based on question $1($b$)$ in the written Assignment $4.$

The program in question $1$ of the written assignment, contains an incomplete recursive function raised$\_$to$\_$power$().$ This function returns the value of the first parameter number of type float raised to the value of the second parameter power of type int for all values of power greater than or equal to $0.$

Based on the fact that any value raised to the power of $0$ is $1,$ the base case for this function is

Select one:

else if $($power $==0)$

return $(1\mathrm{.}0)$;

else if $($power $==1)$

return $(1\mathrm{.}0)$;

else if $($power $>0)$

return $(1\mathrm{.}0)$;

else if $($power $==0)$

return $($number $*$ raised$\_$to$\_$power$($number$,$ power $-1))$;