Question

Помогите пожалуйста))​

Answer 1

Ответ:

$1)\ \ f(x)=x+x^2\ \ ,\ \ \ x_0=-6,4\ ,\ x=-6,5\\\\\Delta x=x-x_0=-6,5+6,4=-0,1\\\\\Delta f=f(x_0+\Delta x)-f(x_0)=f(-6,5)-f(-6,4)=\\\\=(-6,5+42,25)-(-6,4+40,96))=35,75-34,56=1,19\\\\\dfrac{\Delta x}{\Delta f}=\dfrac{-0,1}{1,19}=-\dfrac{10}{119}$

$2)\ \ y=\dfrac{x-1}{x+1}=\ \ ,\ \ x_0=2\\\\\\y(x_0+\Delta x)=\dfrac{2+\Delta x-1}{2+\Delta x+1}=\dfrac{1+\Delta x}{3+\Delta x}\\\\y(x_0)=\dfrac{2-1}{2+1}=\dfrac{1}{3}\\\\\Delta y=\dfrac{1+\Delta x}{3+\Delta x}-\dfrac{1}{3}=\dfrac{3+3\Delta x-3-\Delta x}{3\, (3+\Delta x)}=\dfrac{2\Delta x}{3\, (3+\Delta x)}$

$y'(x_0)=\lim\limits _{\Delta x \to 0}\, \dfrac{\Delta y}{\Delta x}=\lim\limits _{\Delta x \to 0}\, \Big(\dfrac{2\Delta x}{3\, (3+\Delta x)}:\Delta x\Big)=\lim\limits _{\Delta x \to 0}\dfrac{2\Delta x}{3\, (3+\Delta x)\cdot \Delta x}=\\\\\\=\lim\limits _{\Delta x \to 0}\dfrac{2}{3\, (3+\Delta x)}=\dfrac{2}{3\cdot (3+0)}=\dfrac{2}{3\cdot 3}=\dfrac{2}{9}$