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1) \(f\left(x\right)=2x-5\)
\(f'\left(x\right)=2\)
\(\Rightarrow f'\left(4\right)=2\)
2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)
\(\Rightarrow y'=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)
3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)
\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)
1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)
2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)
3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)
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
=>\(\left(x-2\sqrt{2}\right)^2=0\)
=>\(x-2\sqrt{2}=0\)
=>\(x=2\sqrt{2}\)
Lời giải:
\(x\in [-\sqrt{2}; \sqrt{2}]\Rightarrow x^2\leq 2\Rightarrow \sqrt{x^2+1}\leq \sqrt{3}\)
\(y=\frac{x+1}{\sqrt{x^2+1}}\geq \frac{x+1}{\sqrt{3}}\geq \frac{-\sqrt{2}+1}{\sqrt{3}}\)
Vậy $y_{\min}=\frac{-\sqrt{2}+1}{\sqrt{3}}$ khi $x=-\sqrt{2}$
$y^2=\frac{x^2+2x+1}{x^2+1}=1+\frac{2x}{x^2+1}$
$y^2=2+\frac{2x-x^2-1}{x^2+1}=2-\frac{(x-1)^2}{x^2+1}\leq 2$
$\Rightarrow y\leq \sqrt{2}$
Vậy $y_{\max}=\sqrt{2}$ khi $x=1$
1) \(y=\dfrac{2x^2+1}{x^2}\)
\(\Rightarrow y'=\dfrac{\left(4x+1\right)x^2-2x\left(2x^2+1\right)}{x^4}\)
\(\Leftrightarrow y'=\dfrac{4x^3+x^2-4x^3-2x}{x^4}\)
\(\Leftrightarrow y'=\dfrac{x^2-2x}{x^4}=\dfrac{x\left(x-2\right)}{x^4}=\dfrac{x-2}{x^3}\)
2) \(f\left(x\right)=\sqrt[]{-5x^2+14x-9}\)
\(\Rightarrow f'\left(x\right)=\dfrac{-10x+14}{2\sqrt[]{-5x^2+14x-9}}\)
\(\Leftrightarrow f'\left(x\right)=\dfrac{-2\left(5x-7\right)}{2\sqrt[]{-5x^2+14x-9}}\)
\(\Leftrightarrow f'\left(x\right)=\dfrac{-\left(5x-7\right)}{\sqrt[]{-5x^2+14x-9}}\)
Để \(f'\left(x\right)=0\)
\(f'\left(x\right)=\dfrac{-\left(5x-7\right)}{\sqrt[]{-5x^2+14x-9}}=0\)
\(\Leftrightarrow5x-7=0\)
\(\Leftrightarrow5x=7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy tập hợp giá trị để \(f'\left(x\right)=0\) là \(\left\{\dfrac{7}{5}\right\}\)
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
\(f'\left(x\right)=-\dfrac{1}{x^2}\Rightarrow f'\left(\sqrt{2}\right)=-\dfrac{1}{\left(\sqrt{2}\right)^2}=-\dfrac{1}{2}\)
1a.
\(y'=3x^2.f'\left(x^3\right)-2x.g'\left(x^2\right)\)
b.
\(y'=\dfrac{3f^2\left(x\right).f'\left(x\right)+3g^2\left(x\right).g'\left(x\right)}{2\sqrt{f^3\left(x\right)+g^3\left(x\right)}}\)
2.
\(f'\left(x\right)=\left(m-1\right)x^3+\left(m-2\right)x^2-2mx+3=0\)
Để ý rằng tổng hệ số của vế trái bằng 1 nên pt luôn có nghiệm \(x=1\), sử dụng lược đồ Hooc-ne ta phân tích được:
\(\Leftrightarrow\left(x-1\right)\left[\left(m-1\right)x^2+\left(2m-3\right)x-3\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(m-1\right)x^2+\left(2m-3\right)x-3=0\left(1\right)\end{matrix}\right.\)
Xét (1), với \(m=1\Rightarrow x=-3\)
- Với \(m\ne1\Rightarrow\Delta=\left(2m-3\right)^2+12\left(m-1\right)=4m^2-3\)
Nếu \(\left|m\right|< \dfrac{\sqrt{3}}{2}\Rightarrow\) (1) vô nghiệm \(\Rightarrow f'\left(x\right)=0\) có đúng 1 nghiệm
Nếu \(\left|m\right|>\dfrac{\sqrt{3}}{2}\Rightarrow\left(1\right)\) có 2 nghiệm \(\Rightarrow f'\left(x\right)=0\) có 3 nghiệm
1) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)
\(\Rightarrow y=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)
2) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1\left(x+9\right)}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{-6}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(1\right)=\dfrac{-6}{\left(1+3\right)^2}+\dfrac{2}{\sqrt[]{1}}=-\dfrac{3}{8}+2=\dfrac{13}{8}\)