Question

P(x)=4x-x²-3

Tìm GTLN

Answer 1

\(P\left(x\right)=4x-x^2-4+1=-\left(x-2\right)^2+1\le0+1=1\Rightarrow P\left(x\right)_{max}=1\Leftrightarrow x=2\)

Answer 2

\(P\left(x\right)=4x-x^2-3\)

\(=-\left(x^2-4x+3\right)\)

\(=-\left(x^2-4x+4\right)+1\)

\(=-\left(x-2\right)^2+1\le1\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)