Найдите Δx и Δf в точке x₀:
а) f(x)=3x-x², x₀=4,1; x=4,2;
б)f(x)=2x+x²; Х₀=2,3; Х=2,4;
в) f(x)=cosx,x₀=pi/6;x=pi/4;
г) f(x)=tgx,xo=pi/4;x=pi/3.

аΔx=x-x0⇒Δx=4,2-4,1=0,1Δf=f(x)-f(x0)=(3*4,2-4,2²)-(3*4,1-4,1²)=3(4,2-4,1)-(4,2-4,1)(4,2+4,1)==0,3-0,83=-0,53бΔx=2,4-2,3=0,1Δf=2(2,4-2,3)+(2,4²-2,3²)=(2,4-2,3)*(2-(2,4+2,3))=0,1*(-2,7)=-0,27вΔx=π/4-π/6=π/12Δf=cosπ/4-cosπ/6=√2/2-√3/2=(√2-√3)/2гΔx=π/3-π/4=π/12Δf=tgπ/3-tgπ/4=√3-1