To solve for $K$ such that the domain of the function $f(x) = \sqrt{2x + 3|x + 2| + Kx}$ is all real numbers, we need to ensure that the expression inside the square root is always non-negative. This is because the square root function is only defined for non-negative numbers.

Step 1: Analyze the expression inside the square root

The expression inside the square root is: $2x + 3|x + 2| + Kx$ For this to be defined for all $x$, it must always be greater than or equal to zero: $2x + 3|x + 2| + Kx \geq 0$

Step 2: Consider two cases for the absolute value

Case 1: $x + 2 \geq 0$ (i.e., $x \geq -2$)

In this case, $|x + 2| = x + 2$, and the expression becomes: $2x + 3(x + 2) + Kx = 2x + 3x + 6 + Kx = (5 + K)x + 6$ For the domain to include all values of $x \geq -2$, we must ensure that: $(5 + K)x + 6 \geq 0 \text{ for all } x \geq -2$ Let's check this condition.

• When $x = -2$, the expression becomes: $(5 + K)(-2) + 6 = -2(5 + K) + 6 = -10 - 2K + 6 = -4 - 2K$ For this to be non-negative: $-4 - 2K \geq 0 \implies -2K \geq 4 \implies K \leq -2$

Next, consider large values of $x$. For the expression to be non-negative as $x \to \infty$, we require that $5 + K \geq 0$, so: $K \geq -5$

From this case, we get the condition: $-5 \leq K \leq -2$.

Case 2: $x + 2 < 0$ (i.e., $x < -2$)

In this case, $|x + 2| = -(x + 2) = -x - 2$, and the expression becomes: $2x + 3(-x - 2) + Kx = 2x - 3x - 6 + Kx = (-1 + K)x - 6$ For the domain to include $x < -2$, we require that: $(-1 + K)x - 6 \geq 0$ Let's check this condition.

• For large negative values of $x$, the term $(-1 + K)x$ dominates, so for the expression to be non-negative, we need $-1 + K \leq 0$, meaning: $K \leq 1$

Thus, the condition for $x < -2$ becomes $K \leq 1$.

Step 3: Combine the conditions

From Case 1, we have $-5 \leq K \leq -2$, and from Case 2, we have $K \leq 1$. The common solution is: $-5 \leq K \leq -2$

Final Answer:

The value of $K$ must satisfy $-5 \leq K \leq -2$ in order for the domain of the function $\sqrt{2x + 3|x + 2| + Kx}$ to be all real numbers.

Would you like further clarification on any of the steps?

Here are five related questions:

1. How do you solve inequalities involving absolute values?
2. What happens when a function’s domain is restricted by a square root?
3. How do absolute value functions behave in different intervals?
4. Can $K$ take any other values outside the range $[-5, -2]$?
5. How would the graph of this function look like for different values of $K$?

Tip: When dealing with absolute value functions, it’s important to split the function into cases to handle the sign changes properly.