Find the domain and range of the function k(x) = (x+1) / sqrt(2x+1)

Step 1: Find the domain

The function is defined when the expression inside the square root is positive and the denominator is not zero:

$$2x + 1 > 0$$

Solve the inequality:

$$2x > -1$$ $$x > -\frac{1}{2}$$

Domain: $\left( -\frac{1}{2}, \infty \right)$

Step 2: Find the range

Let $y = k(x) = \frac{x+1}{\sqrt{2x+1}}$.

Substitute $t = \sqrt{2x+1}$ where $t > 0$:

$$2x + 1 = t^2$$ $$x = \frac{t^2 - 1}{2}$$

Substitute into $y$:

$$y = \frac{\frac{t^2 - 1}{2} + 1}{t} = \frac{\frac{t^2 - 1 + 2}{2}}{t} = \frac{\frac{t^2 + 1}{2}}{t} = \frac{t^2 + 1}{2t}$$ $$y = \frac{t}{2} + \frac{1}{2t}, \quad t > 0$$

To find the range, analyze $f(t) = \frac{t}{2} + \frac{1}{2t}$ for $t > 0$.

Find the minimum using calculus or AM-GM inequality:

By AM-GM:

$$\frac{t}{2} + \frac{1}{2t} \geq 2\sqrt{\frac{t}{2} \cdot \frac{1}{2t}} = 2\sqrt{\frac{1}{4}} = 1$$

Equality when $\frac{t}{2} = \frac{1}{2t}$, so $t^2 = 1$, $t = 1$ (since $t > 0$).

At $t = 1$, $y = 1$.

As $t \to 0^+$, $y \to \infty$.

As $t \to \infty$, $y \to \infty$.

Thus, $y \geq 1$.

Range: $[1, \infty)$

Final answer

$$\boxed{ \begin{array}{c} \text{Domain: } \left( -\dfrac{1}{2},\ \infty \right) \\[2em] \text{Range: } [1,\ \infty) \end{array} }$$