Answer
Absolute maximum: 15 at x=1 or (1,15)
Absolute minimum: 7 at x=5 or (5,7)
Work Step by Step
f(x)=12+4x-2x^{2} [0,5]
f'(x)=4-4x
0=4-4x
4x=4
x=1
Test with 0:
f'(x)=4-4*0=4 (positive so increasing)
Test with 2:
f'(x)=4-4*2=-4 (negative so decreasing)
So x=1 is a maximum:
f(1)=12+4*1-1^2=15
Test the extremities:
f(0)=12+4*0-0^2=12
f(5)=12+4*5-5^2=7
So x=5 is a minimum.
