task/29412085 -------------------
f'(x₀) = tgα ||α =60° || = tg60° = √3
f'(x) =( (√3 /2)*(x-1)² ) '=(√3 /2)*2*(x-1)*(x-1) ' =√3(x-1)* (x'-1 ')=√3(x-1)* (1 -0)=√3(x-1)
⇒ следовательно f'(x₀) =√3(x₀- 1)
√3(x₀- 1) = √3 ⇔ x₀ - 1 = 1 ⇔ x₀= 1+1 =2.
ответ: 2