A)f(x)=x^2b)f(x)=x^2-3x+2[tex] f(x) = \frac{x - 2}{x ^{2} - 2 } [/tex]f(x)=1/x^3найдите f(-1)-?f(1/2)-?f(2)-?

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a) f(x) = x^2
f(-1) = (-1)^2 = 1
f(1/2) = (1/2)^2 = 1/4
f(2) = 2^2 = 4

b) f(x) = x^2 - 3x + 2
f(-1) = (-1)^2 - 3(-1) + 2 = 1 + 3 + 2 = 6
f(1/2) = (1/2)^2 - 3(1/2) + 2 = 1/4 - 3/2 + 2 = 1/4 - 3/2 + 8/4 = -5/4 + 8/4 = 3/4
f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0

c) f(x) = (x - 2) / (x^2 - 2)
f(-1) = (-1 - 2) / ((-1)^2 - 2) = -3 / (1 - 2) = -3 / -1 = 3
f(1/2) = (1/2 - 2) / ((1/2)^2 - 2) = (-3/2) / (1/4 - 2) = (-3/2) / (1/4 - 8/4) = (-3/2) / (-7/4) = -3/2 * -4/7 = 6/7
f(2) = (2 - 2) / (2^2 - 2) = 0 / (4 - 2) = 0 / 2 = 0

Hence, f(-1) = 3, f(1/2) = 6/7, f(2) = 0.