Đặt \(A=\dfrac{1}{2}x-x^2-4\)
\(=-x^2+\dfrac{1}{2}x-4\)
\(=-\left(x^2-\dfrac{1}{2}x+4\right)\)
\(=-\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{63}{16}\right)\)
\(=-\left[\left(x-\dfrac{1}{4}\right)^2+\dfrac{63}{16}\right]\)
\(=-\left(x-\dfrac{1}{4}\right)^2-\dfrac{63}{16}\)
Ta có: \(-\left(x-\dfrac{1}{4}\right)^2\le0\Rightarrow A=-\left(x-\dfrac{1}{4}\right)^2-\dfrac{63}{16}\le\dfrac{-63}{16}\)
Dấu " = " xảy ra khi \(\left(x-\dfrac{1}{4}\right)^2=0\Rightarrow x=\dfrac{1}{4}\)
Vậy \(MAX_A=\dfrac{-63}{16}\) khi \(x=\dfrac{1}{4}\)
