..第一题b c d问,
1个回答

(a)

9 - x² ≠ 0, domain is x ≠ ±3

Zero: x = 0

(b)

f(x) = |x|(x - 3)/[(3 - x)(3 + x)] = -|x|/(x + 3) (i)

x -> 3, f'(x) -> -3/(3 + 3) = -1/2

(c)

vertical: x = -3

x ≥ 0, f(x) = -x/(x + 3)

f(x)/x = -1/(x + 3)

x -> ∞, f(x)/x -> 0

There're no asymptotes in the form of y = kx + b, where k ≠ 0

x -> ∞: f(x) -> -1

x < 0: f(x) = x/(x + 3)

x ->- ∞, f(x)/x -> 1

horizontal asymptotes: y = -1 and y = 1

(d)

non-removable: x = -3

(x - 3) can be canceled out as shown in (i)

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