If the domain of the function $$f(x) = \sec^{-1}\left(\frac{2x}{5x+3}\right)$$ is $$[\alpha, \beta) \cup (\gamma, \delta]$$, then $$3\alpha + 10\beta + \gamma + 21\delta$$ is equal to _______.

To find the value of the expression $$3\alpha + 10\beta + \gamma + 21\delta$$, we first need to determine the domain of the function $$f(x) = \sec^{-1} \left( \frac{2x}{5x+3} \right)$$.

The domain of $$\sec^{-1}(u)$$ is defined for $$|u| \geq 1$$. This implies:

$$\left| \frac{2x}{5x+3} \right| \geq 1$$

This absolute value inequality splits into two cases:

Case A: $$\frac{2x}{5x+3} \geq 1$$

Case B: $$\frac{2x}{5x+3} \leq -1$$

Subtract 1 from both sides:

$$\frac{2x}{5x+3} - 1 \geq 0 \implies \frac{2x - (5x+3)}{5x+3} \geq 0 \implies \frac{-3x - 3}{5x+3} \geq 0$$

Multiply by $$-1$$ (and flip the inequality sign):

$$\frac{3x+3}{5x+3} \leq 0 \implies \frac{x+1}{x + 3/5} \leq 0$$

Using the wavy curve method, the solution is $$x \in [-1, -3/5)$$.

Add 1 to both sides:

$$\frac{2x}{5x+3} + 1 \leq 0 \implies \frac{2x + (5x+3)}{5x+3} \leq 0 \implies \frac{7x+3}{5x+3} \leq 0$$

The critical points are $$x = -3/7$$ and $$x = -3/5$$.

Using the wavy curve method, the solution is $$x \in (-3/5, -3/7]$$.

The total domain is the union of the two cases:

$$[-1, -3/5) \cup (-3/5, -3/7]$$

Comparing this with the given format $$[\alpha, \beta) \cup (\gamma, \delta]$$, we identify:

• $$\alpha = -1$$
• $$\beta = -3/5$$
• $$\gamma = -3/5$$
• $$\delta = -3/7$$

Now, substitute these values into $$3\alpha + 10\beta + \gamma + 21\delta$$:

$$3(-1) + 10\left(-\frac{3}{5}\right) + \left(-\frac{3}{5}\right) + 21\left(-\frac{3}{7}\right)$$

$$-3 - 6 - \frac{3}{5} - 9$$

Combine the integers:

$$-18 - \frac{3}{5} = -\frac{90}{5} - \frac{3}{5} = -\frac{93}{5}$$