Найдите f(x0), если 1) f(x) =1 /(2x +7)^4 - (1-x) ^3; X0 = - 3 2) f(x) = cos4x +ctgx ; X0 = п/2 3) f(x) = sqrt(корень из)(x^2 - 8x +12) X0 = 4 4) f(x) =xsin (x/3 + п/6) X0 = п

1) f = -4*(2x+7)^-5 *2 -3*(1-x)^2*(-1) = -8/(2x+7)^5 +3*(1-x)^2
f(-3) = -8/(-6+7)^5 +3*(1+3)^2 = -8+3*16 = -8+48 =40

2) f = (-sin4x)*4 -1/sin^2 x= - (4*sin4x +1/sin^2 x)
f(пи/2) = -(4*sin 2пи+ 1/ sin^2 (пи/2)) = -(4*0 +1/1) = -1

3) f = [(x^2-8x+12)^ (1/2)] =  1/2* ((x^2-8x+12)^ (-1/2))*(2x-8) = 2*(x-4)/(2*(x^2-8x+12)^ (1/2))=
=(x-4)/(x^2-8x+12)^ (1/2) = (x-4)/корень(x^2-8x+12)
f(4)=(4-4)/корень(16-32+12)=0

4)f  = sin (x/3+пи/6) +x*cos(x/3+пи/6) * 1/3 = sin (x/3+пи/6)+(x*cos(x/3+пи/6))/3
f(пи)= sin(пи/3+пи/6) + (пи*cos(пи/3+пи/6))/3 = sin (пи/2)+ (пи*cos *(пи/2))/3= 1+ пи*0/3 = 1