1. (1 pt) Let g(y) = 8y2 + 13y + 5. A function f is known to

be continuous on (
−∞, ∞), to satisfy g ◦ f (x) = 0 for all real x
and it is known that the value f (8) is not equal to
−1. You have
enough information to determine f precisely.

8f(x)^2 + 13f(x) + 5 = 0

Than can be factored to give:
(8f(x) + 5)(f(x) + 1) = 0

So either f(x) = -5/8 or -1

But f(x) = -1 has been ruled out for x=8 , so
f(x) = -5/8 for any x.