Question

Найдите Δx и Δf в точке x0:
1) f(x) 3x-x^2, x0=4.1;x=4.2
2) f(x)=2x+x^2, x0=2.3;x=2.4
3)f(x)=cosx, x0=П/6;x=П/4

Answer 1

$f(x)=3x-x^2 \ Delta x=x-x_0=4.2-4.1=0.1 \ Delta y=f(x)-f(x_0)=(3cdot4.2-4.2^2)-(3cdot4.1-4.1^2)= \ =12.6-17.64-12.3+16.81=-0.53$

$f(x)=2x+x^2 \ Delta x=x-x_0=2.4-2.3=0.1 \ Delta y=f(x)-f(x_0)=(2cdot2.4+2.4^2)-(2cdot2.3+2.3^2)= \ =4.8+5.76-4.6-5.29=0.67$

$f(x)=cos x \ Delta x=x-x_0= frac{ pi }{4} - frac{ pi }{6} =frac{ 3pi }{12} - frac{2 pi }{12} = frac{pi }{12} \ Delta y=f(x)-f(x_0)=cos frac{ pi }{4} -cos frac{ pi }{6} =frac{ sqrt{2} }{2} -frac{ sqrt{3} }{2} = frac{ sqrt{2} - sqrt{3} }{2}$