Functions - The inverse function

Given the function $f(x) = 2x - 7$, find the inverse function $f^{-1}(x)$.

1. Let $y = 2x - 7$
2. Swap $x$ and $y$: $x = 2y - 7$
3. Solve for $y$: $x + 7 = 2y$ $y = \frac{x + 7}{2}$
4. Therefore, $f^{-1}(x) = \frac{x + 7}{2}$

We follow the algebraic steps of interchanging the dependent and independent variables and rearranging to make $y$ the subject.

Find the inverse of the rational function $f(x) = \frac{x + 3}{x - 2}$ where $x \neq 2$.

1. Let $y = \frac{x + 3}{x - 2}$
2. Swap $x$ and $y$: $x = \frac{y + 3}{y - 2}$
3. Multiply by $(y - 2)$: $x(y - 2) = y + 3$
4. Expand: $xy - 2x = y + 3$
5. Collect $y$ terms on one side: $xy - y = 2x + 3$
6. Factor out $y$: $y(x - 1) = 2x + 3$
7. Solve for $y$: $y = \frac{2x + 3}{x - 1}$
8. $f^{-1}(x) = \frac{2x + 3}{x - 1}, x \neq 1$

For rational functions, cross-multiplying and factoring the $y$ term is the standard technique to isolate the variable.

If the domain of $f(x) = \sqrt{x - 5}$ is $x \geq 5$ and the range is $y \geq 0$, state the domain and range of $f^{-1}(x)$.

For $f^{-1}(x)$: Domain: $x \geq 0$ Range: $y \geq 5$

The domain and range of a function and its inverse are interchanged. Since the range of $f$ is $[0, \infty)$, that becomes the domain of $f^{-1}$. Since the domain of $f$ is $[5, \infty)$, that becomes the range of $f^{-1}$.

1. Let $y = x^2 + 1$.
2. Interchange $x$ and $y$: $x = y^2 + 1$.
3. Solve for $y$: $y^2 = x - 1 \implies y = \sqrt{x - 1}$ (taking the positive root because the domain of $f$ is $x \geq 0$).
4. Therefore, $f^{-1}(x) = \sqrt{x - 1}$ for $x \geq 1$.

Since the original function is restricted to $x \geq 0$, the inverse is only the positive square root branch. The graph of $f(x)$ starts at $(0,1)$ and moves upwards, while $f^{-1}(x)$ starts at $(1,0)$ and moves to the right.

1. Let $y = e^{x-2}$.
2. Swap $x$ and $y$: $x = e^{y-2}$.
3. Take the natural log of both sides: $\ln(x) = \ln(e^{y-2})$.
4. $\ln(x) = y - 2$.
5. $y = \ln(x) + 2$.
6. $f^{-1}(x) = \ln(x) + 2$. Since the range of $f(x) = e^{x-2}$ is $(0, \infty)$, the domain of $f^{-1}(x)$ is $x > 0$.

The exponential function $e^{x-2}$ is always positive, so its range is $y > 0$. This range becomes the domain for the logarithmic inverse function.