Question

tìm m để hàm số y=x³-3(2m+1)x²+(12+5)x+2 đồng biến trên khoang (2:+\(\infty\)⁾

Answer 1

\(y'=3x^2-6\left(2m+1\right)x+12m+5\ge0\) ;\(\forall x>2\)

\(\Leftrightarrow3x^2-6x+5\ge12m\left(x-1\right)\) ;\(\forall x>2\)

\(\Leftrightarrow\frac{3x^2-6x+5}{12\left(x-1\right)}\ge m\) ;\(\forall x>2\)

Đặt \(g\left(x\right)=\frac{3x^2-6x+5}{12\left(x-1\right)}\Rightarrow m\le\min\limits_{x>2}g\left(x\right)\)

\(g'\left(x\right)=\frac{3x^2-6x+1}{12\left(x-1\right)^2}>0\) ;\(\forall x>2\) \(\Rightarrow g\left(x\right)>g\left(2\right)\) ;\(\forall x>2\)

\(\Rightarrow m\le g\left(2\right)=\frac{5}{12}\)