help

\(y'=x^2-2\left(m-1\right)x+3\left(m-1\right)\)

Hàm đồng biến trên khoảng đã cho khi với mọi \(x>1\) ta luôn có:

\(g\left(x\right)=x^2-2\left(m-1\right)x+3\left(m-1\right)\ge0\)

\(\Rightarrow\min\limits_{x>1}g\left(x\right)\ge0\)

Do \(a=1>0;-\dfrac{b}{2a}=m-1\)

TH1: \(m-1\ge1\Rightarrow m\ge2\)

\(\Rightarrow g\left(x\right)_{min}=f\left(m-1\right)=\left(m-1\right)^2-2\left(m-1\right)^2+3\left(m-1\right)\ge0\)

\(\Rightarrow\left(m-1\right)\left(4-m\right)\ge0\Rightarrow1\le m\le4\Rightarrow2\le m\le4\)

TH2: \(m-1< 1\Rightarrow m< 2\Rightarrow g\left(x\right)_{min}=g\left(1\right)=m\ge0\)

Vậy \(0\le m\le4\)

\(y'=\dfrac{x-m-x+1}{\left(x-m\right)^2}=\dfrac{1-m}{\left(x-m\right)^2}\)

Hàm số nghịch biến trên khoảng \(\left(-\infty;2\right)\Leftrightarrow y'< 0\forall x\in\left(-\infty;2\right)\Leftrightarrow\left\{{}\begin{matrix}1-m< 0\\x\ne m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>1\\m\ge2\end{matrix}\right.\Rightarrow m\ge2\)

Có 19-2+1=18 giá trị nguyên của m thỏa mãn

a: \(y=-x^3-\left(m+1\right)x^2+3\left(m+1\right)x\)

=>\(y'=-3x^2-\left(m+1\right)\cdot2x+3\left(m+1\right)\)

=>\(y'=-3x^2+x\cdot\left(-2m-2\right)+\left(3m+3\right)\)

Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)

=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(-2m-2\right)^2-4\cdot\left(-3\right)\left(3m+3\right)< =0\\-3< 0\end{matrix}\right.\)

=>\(4m^2+8m+4+12\left(3m+3\right)< =0\)

=>\(4m^2+8m+4+36m+36< =0\)

=>\(4m^2+44m+40< =0\)

=>\(m^2+11m+10< =0\)

=>\(\left(m+1\right)\left(m+10\right)< =0\)

TH1: \(\left\{{}\begin{matrix}m+1>=0\\m+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>=-1\\m< =-10\end{matrix}\right.\)

=>\(m\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}m+1< =0\\m+10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< =-1\\m>=-10\end{matrix}\right.\)

=>-10<=m<=-1

b: \(y=-\dfrac{1}{3}x^3+mx^2-\left(2m+3\right)x\)

=>\(y'=-\dfrac{1}{3}\cdot3x^2+m\cdot2x-\left(2m+3\right)\)

=>\(y'=-x^2+2m\cdot x-\left(2m+3\right)\)

Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)

=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-1< 0\\\left(2m\right)^2-4\cdot\left(-1\right)\cdot\left(-2m-3\right)< =0\end{matrix}\right.\)

=>\(4m^2+4\left(-2m-3\right)< =0\)

=>\(m^2-2m-3< =0\)

=>(m-3)(m+1)<=0

TH1: \(\left\{{}\begin{matrix}m-3>=0\\m+1< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>=3\\m< =-1\end{matrix}\right.\)

=>\(m\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}m-3< =0\\m+1>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< =3\\m>=-1\end{matrix}\right.\)

=>-1<=m<=3