Ответ:
1) Найти производные функций .
[tex]\bf f(x)=x^2-2x\ \ ,\ \ f'(x)=2x-2\ \ ,\ \ f'(\dfrac{1}{2})=2\cdot \dfrac{1}{2}-2=1-2=-1\\\\f(x)=x^4-4x\ \ ,\ \ f'(x)=4x^3-4\ \ ,\ \ f'(-1)=-4-4=-8\\\\f(x)=\dfrac{x+1}{2x-3}\ \ ,\ \ f'(x)=\dfrac{2x-3-2\, (x+1)}{(2x-3)^2}=\dfrac{-5}{(2x-3)^2}\ ,\ \ f'(-3)=\dfrac{-5}{81}\\\\f(x)=x-\dfrac{1}{x}\ \ ,\ \ f'(x)=1+\dfrac{1}{x^2}\ \ ,\ \ f'(0,1)=1+\dfrac{1}{0,01}=101[/tex]
2) Уравнение касательной : [tex]\bf y=f(x_0)+f'(x_0)(x-x_0)[/tex] .
[tex]\bf f(x)=\dfrac{1}{6}\, x^3+4x\ \ ,\ \ x_0=-2\\\\f(-2)=\dfrac{1}{6}\cdot (-8)-8=-\dfrac{4}{3}-8=-\dfrac{28}{3}\\\\f'(x)=\dfrac{1}{6}\cdot 3x^2+4=\dfrac{1}{2}\, x^2+4\\\\f'(-2)= \dfrac{1}{2}\cdot 4+4=6\\\\y=-\dfrac{28}{3}+6\, (x+2)\\\\\boxed{\bf \ y=6x+\dfrac{8}{3}\ }[/tex]
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Verified answer
Ответ:
1) Найти производные функций .
[tex]\bf f(x)=x^2-2x\ \ ,\ \ f'(x)=2x-2\ \ ,\ \ f'(\dfrac{1}{2})=2\cdot \dfrac{1}{2}-2=1-2=-1\\\\f(x)=x^4-4x\ \ ,\ \ f'(x)=4x^3-4\ \ ,\ \ f'(-1)=-4-4=-8\\\\f(x)=\dfrac{x+1}{2x-3}\ \ ,\ \ f'(x)=\dfrac{2x-3-2\, (x+1)}{(2x-3)^2}=\dfrac{-5}{(2x-3)^2}\ ,\ \ f'(-3)=\dfrac{-5}{81}\\\\f(x)=x-\dfrac{1}{x}\ \ ,\ \ f'(x)=1+\dfrac{1}{x^2}\ \ ,\ \ f'(0,1)=1+\dfrac{1}{0,01}=101[/tex]
2) Уравнение касательной : [tex]\bf y=f(x_0)+f'(x_0)(x-x_0)[/tex] .
[tex]\bf f(x)=\dfrac{1}{6}\, x^3+4x\ \ ,\ \ x_0=-2\\\\f(-2)=\dfrac{1}{6}\cdot (-8)-8=-\dfrac{4}{3}-8=-\dfrac{28}{3}\\\\f'(x)=\dfrac{1}{6}\cdot 3x^2+4=\dfrac{1}{2}\, x^2+4\\\\f'(-2)= \dfrac{1}{2}\cdot 4+4=6\\\\y=-\dfrac{28}{3}+6\, (x+2)\\\\\boxed{\bf \ y=6x+\dfrac{8}{3}\ }[/tex]