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2: ĐKXĐ: x<>1
\(f'\left(x\right)=\dfrac{\left(x^2-3x+3\right)'\left(x-1\right)-\left(x^2-3x+3\right)\left(x-1\right)'}{\left(x-1\right)^2}\)
\(=\dfrac{\left(2x-3\right)\left(x-1\right)-\left(x^2-3x+3\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2-5x+3-x^2+3x-3}{\left(x-1\right)^2}=\dfrac{x^2-2x}{\left(x-1\right)^2}\)
f'(x)=0
=>x^2-2x=0
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
1:
\(f\left(x\right)=\dfrac{1}{3}x^3-2\sqrt{2}\cdot x^2+8x-1\)
=>\(f'\left(x\right)=\dfrac{1}{3}\cdot3x^2-2\sqrt{2}\cdot2x+8=x^2-4\sqrt{2}\cdot x+8=\left(x-2\sqrt{2}\right)^2\)
f'(x)=0
=>\(\left(x-2\sqrt{2}\right)^2=0\)
=>\(x-2\sqrt{2}=0\)
=>\(x=2\sqrt{2}\)
1. Áp dụng quy tắc L'Hopital
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1}{f\left(0\right)-f\left(x\right)}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2\sqrt{x+1}}}{-f'\left(0\right)}=-\dfrac{1}{6}\)
2.
\(g'\left(x\right)=2x.f'\left(\sqrt{x^2+4}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\f'\left(\sqrt{x^2+4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+4}=1\\\sqrt{x^2+4}=-2\end{matrix}\right.\)
2 pt cuối đều vô nghiệm nên \(g'\left(x\right)=0\) có đúng 1 nghiệm
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{ax+1}-\sqrt[]{1-bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{bx}{1+\sqrt[]{1-bx}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{a}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{b}{1+\sqrt[]{1-bx}}\right)=\dfrac{a}{3}+\dfrac{b}{2}\)
Hàm liên tục tại \(x=0\) khi:
\(\dfrac{a}{3}+\dfrac{b}{2}=3a-5b-1\Leftrightarrow8a-11b=3\)
\(f\left(x\right)=x+\dfrac{\sqrt{x}}{x+1}\Rightarrow f'\left(x\right)=1+\dfrac{1-x}{2\sqrt{x}\left(x+1\right)^2}\)
\(f'\left(x\right)-1>0\Leftrightarrow\dfrac{1-x}{2\sqrt{x}\left(x+1\right)^2}>0\)
\(\Rightarrow0< x< 1\)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(ax+2\right)=a+2\)
Hàm liên tục tại x=1 khi:
\(a+2=\dfrac{1}{4}\Rightarrow a=-\dfrac{7}{4}\)
\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x}{x\left(\sqrt{x+4}+2\right)}=\lim\limits_{x\rightarrow0^+}\dfrac{1}{\sqrt{x+4}+2}=\dfrac{1}{4}\)
\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(mx^2+2m+\dfrac{1}{4}\right)=2m+\dfrac{1}{4}\)
Hàm liên tục tại x=0 khi: \(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=f\left(0\right)\)
\(\Leftrightarrow2m+\dfrac{1}{4}=\dfrac{1}{4}\Leftrightarrow m=0\)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(mx\right)=m\)
Hàm liên tục tại x=1 khi: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)
\(\Leftrightarrow m=\dfrac{1}{4}\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{3\left(x-1\right)}{\left(1-x\right)\left(x^2+x+1\right)\left(\sqrt[3]{\left(3x+5\right)^2}+2\sqrt[3]{3x+5}+4\right)}\)
\(=\lim\limits_{x\rightarrow1^-}\dfrac{-3}{\left(x^2+x+1\right)\left(\sqrt[3]{\left(3x+5\right)^2}+2\sqrt[3]{3x+5}+4\right)}=-\dfrac{1}{12}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{2m\sqrt{x}+3}{5}=\dfrac{2m+3}{5}\)
Hàm liên tục trên R khi và chỉ khi:
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\Leftrightarrow\dfrac{2m+3}{5}=-\dfrac{1}{12}\Leftrightarrow m=-\dfrac{41}{24}\)
\(f\left(0\right)=2.0+m+1=m+1\)
\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[3]{x+1}-1}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x+1-1}{x(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1)}=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)\(f\left(0\right)=\lim\limits_{x\rightarrow0^+}f\left(x\right)\Leftrightarrow m+1=\dfrac{1}{3}\Rightarrow m=-\dfrac{2}{3}\)
\(f'\left(x\right)=x^2-4\sqrt{2}x+8=\left(x-2\sqrt{2}\right)^2\)
\(f'\left(x\right)=0\Rightarrow\left(x-2\sqrt{2}\right)^2=0\Rightarrow x=2\sqrt{2}\)