Question 1209880: Find the all real numbers that are not in the domain of f(g(x)), where
f(x) = \frac{3x^2 - 10x - 25}{x + 1} and g(x) = \frac{14x - 6}{3x^2 + 5x + 15}
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $f(x) = \frac{3x^2 - 10x - 25}{x + 1}$ and $g(x) = \frac{14x - 6}{3x^2 + 5x + 15}$.
We want to find the real numbers that are not in the domain of $f(g(x))$.
First, we find the domain of $f(x)$. The denominator is $x+1$, so $x+1 \neq 0$, which means $x \neq -1$. The domain of $f(x)$ is all real numbers except $x = -1$.
Next, we find the domain of $g(x)$. The denominator is $3x^2 + 5x + 15$. We check the discriminant to see if it has real roots:
$$ \Delta = b^2 - 4ac = 5^2 - 4(3)(15) = 25 - 180 = -155 $$
Since the discriminant is negative, the denominator has no real roots, so $3x^2 + 5x + 15 \neq 0$ for all real numbers $x$.
Thus, the domain of $g(x)$ is all real numbers.
Now, we consider $f(g(x))$. The domain of $f(g(x))$ is all real numbers $x$ such that $g(x)$ is in the domain of $f(x)$.
That is, we need $g(x) \neq -1$.
We need to solve $g(x) = -1$:
$$ \frac{14x - 6}{3x^2 + 5x + 15} = -1 $$
$$ 14x - 6 = -1(3x^2 + 5x + 15) $$
$$ 14x - 6 = -3x^2 - 5x - 15 $$
$$ 3x^2 + 19x + 9 = 0 $$
We can use the quadratic formula to solve for $x$:
$$ x = \frac{-19 \pm \sqrt{19^2 - 4(3)(9)}}{2(3)} = \frac{-19 \pm \sqrt{361 - 108}}{6} = \frac{-19 \pm \sqrt{253}}{6} $$
Thus, we have two values of $x$ that make $g(x) = -1$:
$$ x_1 = \frac{-19 + \sqrt{253}}{6} \quad \text{and} \quad x_2 = \frac{-19 - \sqrt{253}}{6} $$
Since the domain of $g(x)$ is all real numbers, and the only restriction on $f(x)$ is $x \neq -1$, the real numbers not in the domain of $f(g(x))$ are the values of $x$ such that $g(x) = -1$.
Therefore, the real numbers not in the domain of $f(g(x))$ are $\frac{-19 \pm \sqrt{253}}{6}$.
Final Answer: The final answer is $\boxed{\frac{-19 + \sqrt{253}}{6}, \frac{-19 - \sqrt{253}}{6}}$
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