Inverse Functions

Approach:

Follow the steps for finding an inverse function.

Steps:

1. Solve $y = 2x + 1$ for x: $y - 1 = 2x$ $x = \frac{y - 1}{2}$
2. Swap x and y: $y = \frac{x - 1}{2}$
3. Determine the domain: The range of $y = 2x + 1$ is $\mathbb{R}$, so the domain of the inverse is also $\mathbb{R}$.

Answer:

The inverse is $y = \frac{x - 1}{2}$, with domain $\mathbb{R}$.

Approach:

Follow the steps for finding an inverse function, paying attention to the restriction on the domain.

Steps:

1. Solve $y = x^2 + 2$ for x: $y - 2 = x^2$ $x = \sqrt{y - 2}$ (since $x \geq 0$)
2. Swap x and y: $y = \sqrt{x - 2}$
3. Determine the domain: The range of $y = x^2 + 2$ ($x \geq 0$) is $[2, +\infty)$, so the domain of the inverse is $[2, +\infty)$.

Answer:

The inverse is $y = \sqrt{x - 2}$, with domain $[2, +\infty)$.

Approach:

Follow the steps for finding an inverse function.

Steps:

1. Solve $y = \frac{1}{x}$ for x: $xy = 1$ $x = \frac{1}{y}$
2. Swap x and y: $y = \frac{1}{x}$
3. Determine the domain: The range of $y = \frac{1}{x}$ ($x \neq 0$) is $\mathbb{R} \setminus \{0\}$, so the domain of the inverse is also $\mathbb{R} \setminus \{0\}$.

Answer:

The inverse is $y = \frac{1}{x}$, with domain $\mathbb{R} \setminus \{0\}$.

Note: This function is its own inverse; such a function is called a self-inverse function.