The absolute maximum value of the function \(f(x) = 4x-\displaystyle\frac{1}{2}x^2\) in the interval \(\left[-2,\displaystyle\frac{9}{2}\right]\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
On a closed interval, the absolute maximum is the largest of the values at the critical points and the end points.
\(f'(x) = 4 - x = 0,\quad x = 4\)
x = 4 lies in \(\left[-2, \frac{9}{2}\right]\).
\(f(-2) = -8 - 2 = -10\)
\(f(4) = 16 - 8 = 8\)
\( \displaystyle f\left(\frac{9}{2}\right) = 18 - \frac{81}{8} = \frac{63}{8} \)
The largest value is 8, at x = 4.
NCERT (2026-27 edition):
Maximum and Minimum Values of a Function in a Closed Interval
