MCQ
Find the absolute maximum value of $f(x)=4 x-\frac{1}{2} x^2$ in interval $\left[-2, \frac{9}{2}\right]$.
  • 8
  • B
    9
  • C
    6
  • D
    10

Answer

Correct option: A.
8
(A)Here $f(x)=4 x-\frac{1}{2} x^2$
$
\therefore \quad f^{\prime}(x)=4-x
$
now $f^{\prime}(x)=0 \Rightarrow 4-x=0 \Rightarrow x=4$
Thus, $\quad f(-2)=4(-2)-\frac{1}{2}(-2)^2=-8-2=-10$
$\begin{aligned}f(4) & =4(4)-\frac{1}{2}(4)^2=16-8=8 \\\text { and } \quad f\left(\frac{9}{2}\right) & =4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2=18-\frac{81}{8}=\frac{63}{8}\end{aligned}$
Hence absolute maximum value is 8 at $x=4$.

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