AP Calculus problem from 1990 ab exam?

Question

Let f be the function that is given by
f(x)= ax+b/ x^2-c

i) the graph of f is symmetric with respect to the y-axis
ii)lim as x --> 2 + = infinity
iii) f ' (1) = -2

Determine the values of a, b, and c.

if anyone knows please help! I have spent hours and can't figure it out completely!

Answer 1

(i) implies that f(x) = f(-x), so

(ax + b)/(x² - c) = (-ax + b)/(x² - c) for all x

Subtract (ax + b)/(x² - c) from both sides:

0 = (-2ax)/(x² - c) for all x

This forces a = 0

So f(x) = b/(x² - c)

(ii) lim (as x → 2+) f(x) = ∞ means that the denominator of f(x) must go to 0 as x → 2, and that b must be positive. So c = 4.

So f(x) = b/(x² - 4)

f'(x) = -2xb/(x² - 4)²

f'(1) = -2 → -2b/9 = -2 so b = 9 (which is positive, as required for (ii))

Therefore,

f(x) = 9/(x² - 4)