Ответ:
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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Ответ:
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]