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3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
Nhìn thấy đạo hàm bằng định nghĩa là thấy ớn, dài dữ dội
- Khi \(x>1\) \(\Rightarrow f\left(x\right)=\frac{4x-4}{x+1}\)
\(\Delta x=x-x_0\) \(\Rightarrow\Delta y=\frac{4\Delta x+4x_0-4}{x_0+\Delta x+1}-\frac{4x_0-4}{x_0+1}=\frac{8\Delta x}{\left(x_0+1\right)\left(x_0+1+\Delta x\right)}\)
\(\Rightarrow f'\left(x_0\right)=\lim\limits_{\Delta x\rightarrow0}\frac{8\Delta x}{\Delta x\left(x_0+1\right)\left(x_0+1+\Delta x\right)}=\frac{8}{\left(x_0+1\right)^2}\)
- Khi \(x< 1\Rightarrow f\left(x\right)=2x-2\)
\(\Delta x\) là số gia của \(x_0< 1\)
\(\Rightarrow\Delta y=2\left(x_0+\Delta x\right)-2-\left(2x_0-2\right)=2\Delta x\)
\(\Rightarrow f'\left(x_0\right)=\lim\limits_{\Delta x\rightarrow0}\frac{2\Delta x}{\Delta x}=2\)
- Khi \(x\rightarrow1^+\Rightarrow\Delta y\rightarrow2\left(1+\Delta x\right)-2\rightarrow2\Delta x\)
\(\lim\limits_{x\rightarrow1^+}f'\left(x\right)=\lim\limits_{\Delta x\rightarrow0}\frac{2\Delta x}{\Delta x}=2\)
\(\lim\limits_{x\rightarrow1^-}f'\left(x\right)=\lim\limits_{x\rightarrow1^-}\frac{8}{\left(1+1\right)^2}=2\)
\(\Rightarrow f'\left(1\right)=2\)
- Ta có: f ' ( x ) = m - x 2 .
- Do x = -1 là nghiệm của bất phương trình f'(x) < 2.
Chọn B.