For the function $$ h(x)=\frac{3 x}{(x-7)(x+8)} $$, solve the following inequality. $$ h(x) \geq 0 $$

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1. Write the function:
   $$
   h(x)=\frac{3x}{(x-7)(x+8)}
   $$

2. Identify the critical points where either the numerator or denominator is zero:
   - Numerator zero: 
     $$
     3x=0\quad \Longrightarrow \quad x=0.
     $$
   - Denominator zero:
     $$
     x-7=0\quad \Longrightarrow \quad x=7;
     $$
     $$
     x+8=0\quad \Longrightarrow \quad x=-8.
     $$
   The points $$ x=-8 $$ and $$ x=7 $$ are not in the domain.

3. Divide the real line into intervals determined by the points $$ -8 $$, $$ 0 $$, and $$ 7 $$:
   - $$ (-\infty,-8) $$
   - $$ (-8,0) $$
   - $$ (0,7) $$
   - $$ (7,\infty) $$

4. Choose a test point in each interval to determine the sign of $$ h(x) $$.

- For $$ x \in (-\infty,-8) $$ (e.g., $$ x=-10 $$):
     - Numerator: $$ 3(-10)=-30 $$ (negative).
     - Denominator: 
       $$
       (-10-7)(-10+8)=(-17)(-2)=34 \quad \text{(positive)}.
       $$
     - So, $$ h(x) $$ is negative.

- For $$ x \in (-8,0) $$ (e.g., $$ x=-4 $$):
     - Numerator: $$ 3(-4)=-12 $$ (negative).
     - Denominator:
       $$
       (-4-7)(-4+8)=(-11)(4)=-44 \quad \text{(negative)}.
       $$
     - Negative divided by negative yields positive, so $$ h(x) $$ is positive.

- For $$ x \in (0,7) $$ (e.g., $$ x=1 $$):
     - Numerator: $$ 3(1)=3 $$ (positive).
     - Denominator:
       $$
       (1-7)(1+8)=(-6)(9)=-54 \quad \text{(negative)}.
       $$
     - Positive divided by negative yields negative, so $$ h(x) $$ is negative.

- For $$ x \in (7,\infty) $$ (e.g., $$ x=10 $$):
     - Numerator: $$ 3(10)=30 $$ (positive).
     - Denominator:
       $$
       (10-7)(10+8)=(3)(18)=54 \quad \text{(positive)}.
       $$
     - Positive divided by positive is positive, so $$ h(x) $$ is positive.

5. Analyze the equality condition $$ h(x) \geq 0 $$:
   - $$ h(x) $$ is positive for $$ x \in (-8,0) $$ and $$ x \in (7,\infty) $$.
   - The function equals zero when the numerator is zero (and denominator is nonzero), which occurs at $$ x=0 $$. Since $$ x=0 $$ lies in the interval $$ (-8,0) $$, it can be included.

6. Write the solution set, keeping in mind that $$ x=-8 $$ and $$ x=7 $$ are not in the domain:
   $$
   x\in (-8,0] \cup (7,\infty)
   $$

Thus, the solution is:
$$
\boxed{(-8,0] \cup (7,\infty)}
$$
The solution to the inequality $$ h(x) \geq 0 $$ is $$ x $$ in the intervals $$ (-8, 0] $$ and $$ (7, \infty) $$.

For the function \( h(x)=\frac{3 x}{(x-7)(x+8)} \), solve the following inequality. \( h(x) \geq 0 \)

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