Написать уравнение касательной в точке x0=1

a) y=3x^2-2x+5

b) y=4x^3/3-4x

в) y=12x^3-3x^2+x-3

A) f (x) = f (x₀) + f' (x₀) (x-x₀) ;

f (x₀) = 3x₀^2-2x₀+5=3*1²-2*1+5=3-2+5=6;

f' (x) = 6x-2; f' (x₀) = 6x₀-2=6*1-2=4;

f (x) = 6+4 (x-1) = 6+4x-4=2+4x;

б) f (x₀) = 4x₀³/3-4x₀=4*1/3-4=-1/3;

f' (x) = 4x²-4;

f' (x₀) = 4x₀²-4=4*1-4=0

f (x) = f (x₀) + f' (x₀) (x-x₀) = - 1/3+0 * (x-1) = - 1/3;

в) y=12x^3-3x^2+x-3

f (x₀) = 12*1-3*1+1-3=7

f' (x) = 36x²-6x+1;

f' (x₀) = 36*1-6*1+1=31;

f (x) = f (x₀) + f' (x₀) (x-x₀) = 7+31 (x-1) = 7+31x-31=31x-24.