Question

С полным решением пж !!!​

Answer 1

Объяснение:

$y=x^3+x-2\ \ \ \ y=4x-1\\y_k=y(x_0)+y'(x_0)*(x-x_0)\\y(x_0)=x_0^3+x_0-2\\y'=(x^3+x-2)'=3x^2+1\\y'(x_0)=3x_0^2+1\\y_k=x_0^3+x_0-2+(3x_0^2+1)*(x-x_0)=x_0^3+x_0-2+3x_0^2*x+x-3x_0^3-x_0=\\=3x_0^2+x-2x^3_0-2=(3x_0^2+1)*x-2*(x_0^3+1).\\3x_0^2+1=4\\3x_0^2=3\ |:3\\x_0^2=1\\x_0=-1\\b=y_1=-2*((-1)^3+1)=-2*(-1+1)^2=-2*0=0.\\x_0_{'}=1.\\b_{'}=y_0_{'}=-2*(1^3+1)=-2*(1+1)=-2*2=-4.\\y_0+y_0_{'}=0+(-4)=-4.$