在同一坐标系中，画出函数 $y = \ln(2x+9)$ and $y = \sqrt{10-x^2}$ 的图像，标出图像与 $x$ 轴交点的坐标以及所有渐近线的方程。 由此解不等式
$\ln(2x+9) \ge \sqrt{10-x^2}$
并推导不等式 $\ln(2|x|+9) \ge \sqrt{10-x^2}$ 的解。

1. Identify the key features of both graphs to sketch them on the same diagram.

• For $y = \ln(2x+9)$: The domain is $2x+9 > 0 \implies x > -4.5$. Therefore, the vertical asymptote is $x = -4.5$. The $x$-intercept is found where $2x+9 = 1 \implies x = -4$.
• For $y = \sqrt{10-x^2}$: This represents the upper half of a circle centred at the origin with radius $\sqrt{10}$. Its $x$-intercepts are at $(-\sqrt{10}, 0)$ and $(\sqrt{10}, 0)$, and it has a $y$-intercept at $(0, \sqrt{10})$.
  (The student should draw a single coordinate grid plotting these two curves and correctly labelling the intercepts and asymptote).

2. Find the x-coordinates of the points of intersection.

Using a graphing calculator to solve $\ln(2x+9) = \sqrt{10-x^2}$:
$x \approx -2.954 \quad \text{or} \quad x \approx 1.876$

3. Solve the first inequality by identifying where the log graph is above the semi-circle.

We require the region where $y = \ln(2x+9)$ is greater than or equal to $y = \sqrt{10-x^2}$.
From the sketch, this occurs at the outer edges of the semi-circle's domain ($[-\sqrt{10}, \sqrt{10}]$).
Rounding to 3 significant figures where appropriate:
$-\sqrt{10} \le x \le -2.95 \quad \text{or} \quad 1.88 \le x \le \sqrt{10}$

4. Deduce the solution to the second inequality involving a modulus.

The new inequality is $\ln(2|x|+9) \ge \sqrt{10-x^2}$.
Since $x^2 = |x|^2$, we can rewrite the right-hand side as $\sqrt{10-|x|^2}$.
The inequality is exactly the original inequality with $x$ replaced by $|x|$:
$\ln(2|x|+9) \ge \sqrt{10-|x|^2}$
Substitute $|x|$ into the solution from the previous part:
$|x| \in [-\sqrt{10}, -2.954] \cup [1.876, \sqrt{10}]$
Since modulus is non-negative ($|x| \ge 0$):
$1.876 \le |x| \le \sqrt{10}$
Splitting this into the positive and negative cases gives:
$-\sqrt{10} \le x \le -1.88 \quad \text{or} \quad 1.88 \le x \le \sqrt{10}$