‌ Given, ‌‌‌f(x)=4x−‌

1

2

x2
∴‌‌f′(x)=4−‌

1

2

(2x)=4−x
‌ put ‌‌‌f′(x)=0
⇒‌‌4−x=0
⇒‌‌x=4
Then, we evaluate the f at critical point x=4 and at the end points of the interval [−2,‌

9

2

].
f(4)‌=16−‌

1

2

(16)=16−8=8
f(−2)‌=−8−‌

1

2

(4)
‌=−8−2=−10
f(‌

9

2

)‌=4(‌

9

2

)−‌

1

2

(‌

9

2

)2
‌=18−‌

81

8

=7.875
Thus, the absolute maximum value of f on [−2,‌

9

2

] is 8 occurring at x=4.