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f(x) = (ax+b)/(x^2 - c)
i) the graph of f is symmetric about the y-axis
ii) limit as x approaches 2+ of f(x) is positive infinity
iii) f'(1) = -2

Determine the values of a, b, c

I got that c = 4 from the first i). I'm stuck on the second one because
f(x) = f(-x)
(ax+b)/(x^2 - c) = (-ax+b)/(x^2 - c)
ax+b = -ax+b
1=-1 ????

  • Calculus -

    If it is symettric, then a has to be zero. If a is not zero, then it is an odd function. a being zero satisfies your solution.

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