Question 1209882: Find the values of x where the vertical asymptotes of f(g(x)) are located, where
f(x) = \frac{2x - 8}{x^2 - 2x - 3} and g(x) = \frac{x^3 + 2x + 9}{x^2 + 4}.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $f(x) = \frac{2x - 8}{x^2 - 2x - 3}$ and $g(x) = \frac{x^3 + 2x + 9}{x^2 + 4}$.
First, we find the vertical asymptotes of $f(x)$. The denominator of $f(x)$ is $x^2 - 2x - 3 = (x - 3)(x + 1)$. Thus, $f(x)$ has vertical asymptotes at $x = 3$ and $x = -1$.
Now, we want to find the vertical asymptotes of $f(g(x))$. This occurs when $g(x)$ is equal to the values where $f(x)$ has vertical asymptotes, i.e., $g(x) = 3$ and $g(x) = -1$.
1. **$g(x) = 3$**
$$ \frac{x^3 + 2x + 9}{x^2 + 4} = 3 $$
$$ x^3 + 2x + 9 = 3(x^2 + 4) $$
$$ x^3 + 2x + 9 = 3x^2 + 12 $$
$$ x^3 - 3x^2 + 2x - 3 = 0 $$
We can factor this by grouping:
$$ x^2(x - 3) + 1(2x - 3) \neq 0$$
Let's try to factor $x^3 - 3x^2 + 2x - 3 = 0$. Using the rational root theorem, we can't find any rational roots.
Let $h(x) = x^3 - 3x^2 + 2x - 3$. Then $h(3) = 27 - 27 + 6 - 3 = 3 > 0$, and $h(2) = 8 - 12 + 4 - 3 = -3 < 0$. So there is a root between 2 and 3. By using numerical method, we find the real root is approximately $x \approx 2.723$.
2. **$g(x) = -1$**
$$ \frac{x^3 + 2x + 9}{x^2 + 4} = -1 $$
$$ x^3 + 2x + 9 = -(x^2 + 4) $$
$$ x^3 + 2x + 9 = -x^2 - 4 $$
$$ x^3 + x^2 + 2x + 13 = 0 $$
Let $k(x) = x^3 + x^2 + 2x + 13$.
$k(-2) = -8 + 4 - 4 + 13 = 5 > 0$
$k(-3) = -27 + 9 - 6 + 13 = -11 < 0$
So there is a root between -3 and -2. By using numerical method, we find the real root is approximately $x \approx -2.205$.
The denominator of $g(x)$, $x^2 + 4$, is always positive, so $g(x)$ has no vertical asymptotes.
Therefore, the vertical asymptotes of $f(g(x))$ occur at the values of $x$ that solve $g(x) = 3$ and $g(x) = -1$, which are approximately $x \approx 2.723$ and $x \approx -2.205$.
Final Answer: The final answer is $\boxed{2.723, -2.205}$ (approximate values).
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