f(x)=2-x^2
x0=3
y=f(x0)+ f '(x0)(x-x0)
f(3) = 2-9 = -7
f '(x)= (2-x^2)' = -2x
f '(3)= -2*3 = -6
y= -7-6(x-3) = -6x+18-7= -6x+11
y(-2,5) = (-6)*(-2,5)+11 = 26
ОТВЕТ: 26
-(2х+1)<3(x-2) -2x+1<3x-6 -2x-3x<-6-1 -5x<-5 x>5:5 x>1
Вроде бы так
f(x)=2-x^2
x0=3
y=f(x0)+ f '(x0)(x-x0)
f(3) = 2-9 = -7
f '(x)= (2-x^2)' = -2x
f '(3)= -2*3 = -6
y= -7-6(x-3) = -6x+18-7= -6x+11
y(-2,5) = (-6)*(-2,5)+11 = 26
ОТВЕТ: 26