Помогитее 212 поожалуйстаа. Created by ritalednyakova. algebra-ru

Помогитее 212 поожалуйстаа...

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Помогитее 212 поожалуйстаа

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212.
а) f(x) = x² - 3x
f'(x) = 2x - 3
f'(-1/2) = 2·(-1/2) - 3 = -1 - 3 = -4
f'(2) = 2·2 - 3 = 4 - 3 = 1

б) f(x) = x - 4√x
f'(x) = 1 - 4/2√x = 1 - 2/√x
f'(0,01) = 1 - 2/√0,01 = 1 - 2/0,1 = 1 - 20 = -19
f'(4) = 1 - 2/√4 = 1 - 2/2 = 1 - 1 = 0

в) f(x) = x - 1/x
f'(x) = 1 + 1/x²
f'(√2) = 1 + 1/(√2)² = 1 + 1/2 = 3/2 = 1,5
f'(-1/√3) = 1 + 1/(-1/√3)² = 1 + 1/(1/3) = 1 + 3 = 4

г) f(x) = (3 - x)/(2 + x)
f'(x) = [(3 - x)'(2 + x) - (2 + x)'(3 - x)]/(2 + x)² = (-2 - x - 3 + x)/(2 + x)² =
-5/(2 + x)²
f'(-3) = -5/(2 - 3)² = -5/1 = -5
f'(0) = -5/(0 + 2)² = -5/4 = -1,25

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amin07am

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