Question

Tìm GTLN của hàm số \(y=\left(x+2\right)\left(3-x\right)\), với \(-2\le x\le3\).

Answer 1

\(y=\left(x+2\right)\left(3-x\right)\)

\(=3x-x^2+6-2x\)

\(=-x^2+x+6\)

=>y'=-2x+1

Đặt y'=0

=>-2x+1=0

=>-2x=-1

=>\(x=\dfrac{1}{2}\)

\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+2\right)\left(3-\dfrac{1}{2}\right)=\dfrac{5}{2}\cdot\dfrac{5}{2}=\dfrac{25}{4}\)

\(f\left(-2\right)=\left(-2+2\right)\left(3+2\right)=0\)

\(f\left(3\right)=\left(3+2\right)\left(3-3\right)=0\)

=>\(y_{max\left[-2;3\right]}=\dfrac{25}{4}\)