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\(y'=x^2-2x+m\)
\(y'\ge0\) ; \(\forall x\in\left(1;3\right)\Leftrightarrow x^2-2x+m\ge0\) ;\(\forall x\in\left(1;3\right)\)
\(\Leftrightarrow m\ge\max\limits_{\left(1;3\right)}\left(-x^2+2x\right)\)
Xét hàm \(f\left(x\right)=-x^2+2x\) trên \(\left(1;3\right)\)
\(-\dfrac{b}{2a}=1\) ; \(f\left(1\right)=1\) ; \(f\left(3\right)=-3\)
\(\Rightarrow m\ge1\)
\(y'=4mx^3+2mx=2mx\left(2x^2+1\right)\)
Do \(2x\left(x^2+1\right)>0\) ;\(\forall x>0\)
\(\Rightarrow y'\ge0\) ;\(\forall x>0\) khi và chỉ khi \(m>0\)
\(y'=-3x^2-6x+m\Rightarrow y''=-6x-6\)
\(y''=0\Leftrightarrow-6x-6=0\Leftrightarrow x=-1\notin\left[0;1\right]\)
\(\left\{{}\begin{matrix}y'\left(0\right)=m\\y'\left(1\right)=m-9\end{matrix}\right.\Rightarrow^{max}_{\left[0;1\right]}y'=y'\left(0\right)=m\)
\(\Rightarrow m=10\)
`f'(x) = x^2 - 4x+m`
`f'(x) >=0 <=>x^2-4x+m>=0`
`<=> \Delta' >=0`
`<=> 2^2-1.m>=0`
`<=> m<=4`
Vậy....
\(y'=\left(m+1\right)x^2-2\left(m+1\right)x-m\)
\(m=-1\Rightarrow y'=1>0\forall x\in R\)
\(m\ne-1\Rightarrow y'>0\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\\Delta'< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-1\\\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}< 0\left(vl\right)\end{matrix}\right.\)
Vậy với m=-1 thì...
\(y'=\dfrac{\left(2x-m\right)\left(x^2+1\right)-2x\left(x^2-mx+m\right)}{\left(x^2+1\right)^2}=\dfrac{2x-mx^2-m+2mx^2-2mx}{\left(x^2+1\right)^2}=\dfrac{mx^2+2\left(1-m\right)x-m}{\left(x^2+1\right)^2}\)
\(y'=0\Leftrightarrow mx^2+2\left(1-m\right)x-m=0\)
Xet \(m=0\) ko thoa man pt
Xet \(m\ne0\)
\(\left\{{}\begin{matrix}\Delta'>0\\\dfrac{2\left(m-1\right)}{m}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(1-m\right)^2+m^2>0\left(ld\right)\\m=-2\end{matrix}\right.\Rightarrow m=-2\)
(C) : \(y=\dfrac{2x^2+mx+m}{x+1}=\dfrac{2x^2}{x+1}+m\Rightarrow y'=2\left(\dfrac{x^2}{x+1}\right)'\) =
\(2.\dfrac{2x\left(x+1\right)-x^2}{\left(x+1\right)^2}=2.\dfrac{x^2+2x}{\left(x+1\right)^2}=\dfrac{2x^2+4x}{\left(x+1\right)^2}\)
G/s d là tiếp tuyến của (C) tại M(xo ; yo)
PTTT d : \(y=\dfrac{2x_o^2+4x_o}{\left(x_o+1\right)^2}\left(x-x_o\right)+\dfrac{2x_o^2+mx_o+m}{x_o+1}\)
Suy ra : d đi qua A(0;1) nên : \(1=\dfrac{2x_o^2+4x_o}{\left(x_o+1\right)^2}.\left(-x_o\right)+\dfrac{2x_o^2+mx_o+m}{\left(x_o+1\right)}\)
\(\Leftrightarrow-2x_o^3-4x_o^2+2x_o^3+mx_o^2+mx_o+2x_o^2+mx_o+m-\left(x_o+1\right)^2=0\)
\(\Leftrightarrow\left(m-3\right)x_o^2+2\left(m-1\right)x_o+m-1=0\) (*)
m = 3 t/m
m khác 3 . (*) có no \(\Leftrightarrow\Delta'\ge0\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m-3\right)\ge0\)
\(\Leftrightarrow2\left(m-1\right)\ge0\Leftrightarrow m\ge1\)
=> m \(\ge1\) thì từ A(0;1) kẻ được bất kì tiếp tuyến nào đến (C)
https://hoc24.vn/cau-hoi/tim-dong-bien-nghich-bien-cua-ham-so-saua-yleftx2-4x3right4x3b-yleftx2-2x-3right.6358138837048
Giúp mik vs ạ
\(y'=\dfrac{-m^2-1}{\left(x-m\right)^2}\)
\(y'< 0\) ;\(\forall x\in\left(0;1\right)\Leftrightarrow\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\)