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