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Ответ:

Найти отношение приращения функции к приращению аргумента .

[tex]\bf 1)\ \ f(x)=3x^2+1\\\\\Delta f(x)=f(x+\Delta x)-f(x)\\\\\Delta f(x)=3(x+\Delta x)^2+1-(3x^2+1)=\\\\=3x^2+6x\cdot \Delta x+3(\Delta x)^2+1-3x^2-1=6x\cdot \Delta x+3(\Delta x)^2\\\\\dfrac{\Delta f(x)}{\Delta x}=6x+3\cdot \Delta x[/tex]  

[tex]\bf 2)\ \ f(x)=x^2-2x\\\\\Delta f(x)=(x+\Delta x)^2-2(x+\Delta x)-(x^2-2x)=\\\\=x^2+2x\cdot \Delta x+(\Delta x)^2-2x-2\Delta x-x^2+2x=2x\cdot \Delta x+(\Delta x)^2-2\Delta x\\\\\dfrac{\Delta f(x)}{\Delta x}=2x+\Delta x-2[/tex]    

[tex]\bf 3)\ \ f(x)=\dfrac{1}{x}\\\\\Delta f(x)=\dfrac{1}{x+\Delta x}-\dfrac{1}{x}=\dfrac{x-x-\Delta x}{x(x+\Delta x)}=-\dfrac{\Delta x}{x(x+\Delta x)}\\\\\dfrac{\Delta f(x)}{\Delta x}=-\dfrac{1}{x(x+\Delta x)}=-\dfrac{1}{x^2+x\, \Delta x}[/tex]  

[tex]\bf 4)\ \ f(x)=\sqrt{3x}\\\\\Delta f(x)=\sqrt{3(x+\Delta x)}-\sqrt{3x}\\\\\dfrac{\Delta f(x)}{\Delta x}=\dfrac{\sqrt{3(x+\Delta x)}-\sqrt{3x}}{\Delta x}=\dfrac{3x+3\cdot \Delta x-3x}{\Delta x\cdot (\sqrt{3(x+\Delta x)}+\sqrt{3x})}=\\\\=\dfrac{3}{\sqrt{3x+3\, \Delta x}+\sqrt{3x}}[/tex]