The correct option is A 21 and 1
First, isolate the possible values of x at which maximum and minimum value of function possible which is x=0 and x=5.
Now find out the critical points by using
f′(x)=0
f′(x)=3x2−12x+9=0
3(x−1)(x−3)=0
x=1,3
So to find out the maximum and minimum value, just calculate the value of function at x=0,1,3,5 and compare.
x0135f(x)15121
Hence, maximum value is 21 and minimum value is 1.