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.

А
Δx=x-x0⇒Δx=4,2-4,1=0,1
Δf=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=-0,53
б
Δx=2,4-2,3=0,1
Δf=2(2,4-2,3)+(2,4²-2,3²)=(2,4-2,3)*(2-(2,4+2,3))=0,1*(-2,7)=-0,27
в
Δx=π/4-π/6=π/12
Δf=cosπ/4-cosπ/6=√2/2-√3/2=(√2-√3)/2
г
Δx=π/3-π/4=π/12
Δf=tgπ/3-tgπ/4=√3-1