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Надежда199
@Надежда199
August 2022
1
4
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Найдите значение функций:
а) f(x)=4x³+x²+1 в точке -2;
б) f(x)=3/x²+x в точке 1/2;
в) f(x)=1/x+1-x³ в точке 1/3.
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dianah4
А) f(-2) = 4*(-2)³ + (-2)² + 1 = 4*(-8) + 4 + 1 = - 32 + 4 + 1 = -27
б) f(1/2) = 3/(1/2)² + 1/2 = 3/1/4 + 1/2 = 3*4 + 1/2 = 12 + 1/2 =
в) f(1/3) = 1/1/3 + 1 - (1/3)³ = 3 + 1 - 1/27 = 4 - 1/27 =
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aburibaeva
f(-2) = 4*(-2)³ + (-2)² + 1 = 4*(-8) + 4 + 1 = - 32 + 4 + 1 = -27
б) f(1/2) = 3/(1/2)² + 1/2 = 3/1/4 + 1/2 = 3*4 + 1/2 = 12 + 1/2 = 12\frac{1}{2}
в) f(1/3) = 1/1/3 + 1 - (1/3)³ = 3 + 1 - 1/27 = 4 - 1/27 = 3\frac{26}{27}
aburibaeva
ввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввввв
dianah4
и зачем это?
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Answers & Comments
б) f(1/2) = 3/(1/2)² + 1/2 = 3/1/4 + 1/2 = 3*4 + 1/2 = 12 + 1/2 =
в) f(1/3) = 1/1/3 + 1 - (1/3)³ = 3 + 1 - 1/27 = 4 - 1/27 =
б) f(1/2) = 3/(1/2)² + 1/2 = 3/1/4 + 1/2 = 3*4 + 1/2 = 12 + 1/2 = 12\frac{1}{2}
в) f(1/3) = 1/1/3 + 1 - (1/3)³ = 3 + 1 - 1/27 = 4 - 1/27 = 3\frac{26}{27}