Уравнение касательной
[tex]\displaystyle y=f(x_0)+f`(x_0)(x-x_0)\\\\f(x)=\frac{x-2}{x+2}+\frac{x+2}{x-2}=\frac{(x-2)^2+(x+2)^2}{x^2-4}=\frac{x^2-4x+4+x^2+4x+4}{x^2-4}=\\\\=\frac{2x^2+8}{x^2-4}\\\\f`(x)=\frac{4x(x^2-4)-(2x^2+8)(2x)}{(x^2-4)^2}=\frac{4x^3-16x-4x^3-16x}{(x^2-4)^2}=\frac{-32x}{(x^2-4)^2}\\\\x_0=3\\\\f(3)=\frac{2*9+8}{9-4}=\frac{26}{5}=5.2\\\\f`(3)=\frac{-32*3}{(9-4)^2}=\frac{-96}{25}=-3.84\\\\y=5.2-3.84(x-3)=5.2-3.84x+11.52=16.72-3.84x[/tex]
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Уравнение касательной
[tex]\displaystyle y=f(x_0)+f`(x_0)(x-x_0)\\\\f(x)=\frac{x-2}{x+2}+\frac{x+2}{x-2}=\frac{(x-2)^2+(x+2)^2}{x^2-4}=\frac{x^2-4x+4+x^2+4x+4}{x^2-4}=\\\\=\frac{2x^2+8}{x^2-4}\\\\f`(x)=\frac{4x(x^2-4)-(2x^2+8)(2x)}{(x^2-4)^2}=\frac{4x^3-16x-4x^3-16x}{(x^2-4)^2}=\frac{-32x}{(x^2-4)^2}\\\\x_0=3\\\\f(3)=\frac{2*9+8}{9-4}=\frac{26}{5}=5.2\\\\f`(3)=\frac{-32*3}{(9-4)^2}=\frac{-96}{25}=-3.84\\\\y=5.2-3.84(x-3)=5.2-3.84x+11.52=16.72-3.84x[/tex]
номер 2 задания нет. Условия задания не написано