Question

tìm m để y'≥0 với mọi x

y=4x^3-(m+1)x^2-mx+4

 

Answer 1

Ta có: \(y'=12x^2-2\left(m+1\right)x-m\)

Để \(y'\ge0\) thì \(12x^2-2\left(m+1\right)x-m\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(m+1\right)^2-12.\left(-m\right)\le0\\12>0\end{matrix}\right.\)

\(\Rightarrow m^2+14m+1\le0\)

\(\Leftrightarrow-7-4\sqrt{3}\le m\le-7+4\sqrt{3}\)

Vậy...

Answer 2

\(y'=12x^2-2\left(m+1\right)x-m\)

\(y'\ge0\) ; \(\forall x\Leftrightarrow12x^2-2\left(m+1\right)x-m\ge0\) ;\(\forall x\)

\(\Leftrightarrow\Delta'=\left(m+1\right)^2+12m\le0\)

\(\Leftrightarrow m^2+14m+1\le0\)

\(\Rightarrow-7-4\sqrt{3}\le m\le-7+4\sqrt{3}\)