Question

ПОМОГИТЕ С 3 ЗАДАНИЯМИ!!!!!

Answer 1

$7)\ \ f(x)=\sqrt{x}\ \ ,\ \ x_0=0,01\\\\k=f'(x_0)\\\\f'(x)=\dfrac{1}{2\sqrt{x}}\ \ ,\ \ \ k=f'(0,01)=\dfrac{1}{2\sqrt{0,01}}=\dfrac{1}{2\cdot 0,1}=\dfrac{1}{0,2}=5\\\\\\Otvet:\ \ k=5\ .$

$8)\ \ f(x)=\dfrac{x^3+3}{\sqrt{x}}\\\\\\f'(x)=\dfrac{3x^2\cdot \sqrt{x}-(x^3+3)\cdot \frac{1}{2\sqrt{x}}}{x}=\dfrac{3x^{\frac{5}{2}}-\frac{1}{2}\, x^{\frac{5}{2}}+\frac{3}{2}\, x^{-\frac{1}{2}}}{x}=\dfrac{\frac{5}{2}\, x^{\frac{5}{2}}+\frac{3}{2}\, x^{-\frac{1}{2}}}{x}=0\\\\\\\dfrac{5}{2}\, x^{\frac{5}{2}}+\dfrac{3}{2}\, x^{-\frac{1}{2}}=0\ ,\ x\ne 0\\\\\\\dfrac{3}{2}\, x^{-\frac{1}{2}}\cdot \Big(\dfrac{5}{3}\, x^3+1\Big)=0$

$\dfrac{5}{3}\, x^3+1=0\ \ ,\ \ x^3=-\dfrac{3}{5}\ \ ,\ \ \ x=-\sqrt[3]{\dfrac{3}{5}}$

$9)\ \ \ f(x)=-\dfrac{x^4}{27}+\dfrac{x^2}{3}-2x+5\ \ ,\ \ x_0=3\\\\\\f(3)=-\dfrac{81}{27}+\dfrac{9}{3}-6+5=-3+3-6+5=-1\\\\f'(x)=-\dfrac{4x^3}{27}+\dfrac{2x}{3}-2\ \ ,\ \ \ f'(3)=-4+2-2=-4\\\\y=f(x_0)+f'(x_0)(x-x_0)\\\\y=-1-4(x-3)\\\\\underline {\ y=-4x+11\ }$