To find the domain of the function $g(x) = \log(x + 4x^2 - 9)$, we need to determine when the argument of the logarithm is positive:

$x + 4x^2 - 9 > 0$

This can be rearranged as:

$4x^2 + x - 9 > 0$

Next, we will find the roots of the quadratic equation $4x^2 + x - 9 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 4$, $b = 1$, and $c = -9$:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot (-9)}}{2 \cdot 4}$ $x = \frac{-1 \pm \sqrt{1 + 144}}{8}$ $x = \frac{-1 \pm \sqrt{145}}{8}$

Calculating the roots:

$x_1 = \frac{-1 + \sqrt{145}}{8}, \quad x_2 = \frac{-1 - \sqrt{145}}{8}$

Now, let's evaluate the approximate values of the roots:

• $\sqrt{145} \approx 12.04$
• Thus, $x_1 \approx \frac{11.04}{8} \approx 1.38$
• And $x_2 \approx \frac{-13.04}{8} \approx -1.63$

Now we have two roots: $x_1 \approx 1.38$ and $x_2 \approx -1.63$.

Next, we need to test the intervals defined by these roots: $(-\infty, x_2)$, $(x_2, x_1)$, and $(x_1, \infty)$.

1. Interval $(-\infty, -1.63)$: Choose $x = -2$: $4(-2)^2 + (-2) - 9 = 16 - 2 - 9 = 5 > 0$

2. Interval $(-1.63, 1.38)$: Choose $x = 0$: $4(0)^2 + (0) - 9 = -9 < 0$

3. Interval $(1.38, \infty)$: Choose $x = 2$: $4(2)^2 + (2) - 9 = 16 + 2 - 9 = 9 > 0$

From this analysis, we find that the function is positive in the intervals:

• $(-\infty, -1.63)$
• $(1.38, \infty)$

Thus, the domain of $g(x)$ in interval notation is:

$(-\infty, \frac{-1 - \sqrt{145}}{8}) \cup (\frac{-1 + \sqrt{145}}{8}, \infty)$

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. What is the range of the function $g(x)$?
2. How do you find the critical points of $g(x)$?
3. What is the behavior of $g(x)$ at the boundaries of the domain?
4. How can you graph the function $g(x)$?
5. What happens to $g(x)$ as $x$ approaches the roots?

Tip: Always check if the argument of a logarithmic function is positive to determine its domain!