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

[latex]f(x)=3x-x^2 \\\ \Delta x=x-x_0=4.2-4.1=0.1 \\\ \Delta y=f(x)-f(x_0)=(3\cdot4.2-4.2^2)-(3\cdot4.1-4.1^2)= \\\ =12.6-17.64-12.3+16.81=-0.53[/latex] [latex]f(x)=2x+x^2 \\\ \Delta x=x-x_0=2.4-2.3=0.1 \\\ \Delta y=f(x)-f(x_0)=(2\cdot2.4+2.4^2)-(2\cdot2.3+2.3^2)= \\\ =4.8+5.76-4.6-5.29=0.67[/latex] [latex]f(x)=\cos x \\\ \Delta x=x-x_0= \frac{ \pi }{4} - \frac{ \pi }{6} =\frac{ 3\pi }{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} [/latex]