Найдите стационарные точки функции f(x)=-⅓x³-½x²+2x+4 в точке с абциссой х=-1
f(x)=-⅓x³-½x²+2x+4
f'(x)=-x^2-x+2=0
D=1-4*-1*2=9
x=1+3/-2 =-2
x2=1-3/-2 = 1
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f(x)=-⅓x³-½x²+2x+4
f'(x)=-x^2-x+2=0
D=1-4*-1*2=9
x=1+3/-2 =-2
x2=1-3/-2 = 1