Question #54f6a

Jan 8, 2017

c. $f \left(x\right) = {x}^{3} - 5 {x}^{2} - 9 x + 45$, $g \left(x\right) = 0$

Explanation:

a. $0 \in \left(- 3 , 3\right)$ and $f \left(0\right) = - 45 < 0 = g \left(0\right)$. As $0 \in \left(- 3 , 3\right)$. Eliminate this choice.

b. When $x$ grows large, a polynomial function begins to resemble its greatest degree term. In this case, this means $g \left(x\right)$ resembles ${x}^{3}$ as $x$ grows large. Take any large $x$ (say, ${10}^{100}$). ${10}^{100} \in \left(5 , \infty\right)$ and $g \left({10}^{100}\right) > 0 = f \left({10}^{100}\right)$. Eliminate this choice.

d. Similar to the above, the different constant term doesn't matter. We can make $g \left(x\right)$ greater than $0$ by choosing a large $x$, and so it will be greater than $f \left(x\right)$ for at least some values in $\left(5 , \infty\right)$. Eliminate this choice.

By process of elimination, the only remaining choice is c., but let's prove it to be the case.

Let $f \left(x\right) = {x}^{3} - 5 {x}^{2} - 9 x + 45$ and $g \left(x\right) = 0$. Factoring $f \left(x\right)$, we get

$f \left(x\right) = \left(x + 3\right) \left(x - 3\right) \left(x - 5\right)$

So $f \left(x\right) = 0$ for $x \in \left\{- 3 , 3 , 5\right\}$. If we partition the reals at these points, we can test to see what sign $f \left(x\right)$ will take on in each of the resulting intervals.

As a shortcut, we know by looking at the first ${x}^{3}$ term that $f \left(x\right) < 0$ on $\left(- \infty , - 3\right)$ and $x > 0$ on $\left(5 , \infty\right)$.

Setting $x = 0$ to test $\left(- 3 , 3\right)$, we find $f \left(0\right) = 45 > 0$, meaning $f \left(x\right) > 0$ on that interval.

Setting $x = 4$ we find that $f \left(x\right) = \left(4 + 3\right) \left(4 - 3\right) \left(4 - 5\right) = - 7 < 0$, meaning $f \left(x\right) < 0$ on that intervals.

Thus $f \left(x\right) > 0 = g \left(x\right)$ for $x \in \left(- 3 , 3\right) \cup \left(5 , \infty\right)$, as desired.