Tìm giá trị lớn nhất cuả đa thức
f(x)= −x2 −2x+2017
Ta có : x2 \(\ge0\Rightarrow-x^2\le0\)
-2x \(\le0\)
=> −x2 −2x+2017 \(\le2017\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=0\end{matrix}\right.\)
Vậy f(x) đạt GTLN bằng 2017 khi x=0
\(f\left(x\right)=-x^2-2x+2017\)
\(=-x^2-x-x-1+2018\)
\(=-x\left(x+1\right)-\left(x+1\right)+2018\)
\(=\left(-x-1\right)\left(x+1\right)+2018\)
\(=-\left(x+1\right)\left(x+1\right)+2018\)
\(=-\left(x+1\right)^2+2018\)
Ta có: \(-\left(x+1\right)^2\le0\)
\(\Rightarrow-\left(x+1\right)^2+2018\le2018\)
Dấu " = " khi \(-\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy \(MAX_{f\left(x\right)}=2018\) khi x = -1
