a. 0∈(−3,3) and f(0)=−45<0=g(0). As 0∈(−3,3). Eliminate this choice.
b. When x grows large, a polynomial function begins to resemble its greatest degree term. In this case, this means g(x) resembles x3 as x grows large. Take any large x (say, 10100). 10100∈(5,∞) and g(10100)>0=f(10100). Eliminate this choice.
d. Similar to the above, the different constant term doesn't matter. We can make g(x) greater than 0 by choosing a large x, and so it will be greater than f(x) for at least some values in (5,∞). Eliminate this choice.
By process of elimination, the only remaining choice is c., but let's prove it to be the case.
Let f(x)=x3−5x2−9x+45 and g(x)=0. Factoring f(x), we get
f(x)=(x+3)(x−3)(x−5)
So f(x)=0 for x∈{−3,3,5}. If we partition the reals at these points, we can test to see what sign f(x) will take on in each of the resulting intervals.
As a shortcut, we know by looking at the first x3 term that f(x)<0 on (−∞,−3) and x>0 on (5,∞).
Setting x=0 to test (−3,3), we find f(0)=45>0, meaning f(x)>0 on that interval.
Setting x=4 we find that f(x)=(4+3)(4−3)(4−5)=−7<0, meaning f(x)<0 on that intervals.
Thus f(x)>0=g(x) for x∈(−3,3)∪(5,∞), as desired.
