Consider the function $f(x) = x^2$.

Note that $f(-4) = (-4)^2 = 16 = 4^2 = f(4)$. In this case, we see that the input values $-4$ and $4$ are different, but the outputs $f(-4) = f(4)$ are the same.

In other words, two different input values are sent to the same output value. If a function always sends two different input values to two different output values, we call this phenomena **one-to-one.**

A function $f(x)$ with domain $A$ is called **one-to-one** if no two elements of $A$ are sent to the same value.

In other words, if $x_1\neq x_2$, then it is always the case that $f(x_1) \neq f(x_2)$.

In other words, if $x_1\neq x_2$, then it is always the case that $f(x_1) \neq f(x_2)$.

In this class, there are two ways of determining whether a function is one-to-one: algebraically and by using the graph. We will focus on the graphical method.

Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects the graph of $f(x)$ more than once.

Let's look at a few examples on how to use the horizontal line test.

Are the functions \[f(x) = \sqrt{x}\qquad g(x) = x^2 \qquad h(x) = \frac{1}{x}\] one-to-one?

Play around with the horizontal line and see which ones do not pass the horizontal line test.
We see that $f(x)$ and $h(x)$ are one-to-one while $g(x)$ is not.

Let $f$ be a one-to-one function with domain $A$ and range $B$. Then the inverse function, denoted $f^{-1}$, has domain $B$ and range $A$ and is defined by \[f^{-1}(y) = x \text{ if and only if } f(x) = y\]

Notes:

- Looking at the definition above, the inverse function $f^{-1}$ plugs in the $y$ value, and spits back out the original $x$ value. This is why $f^{-1}$ is called an inverse: it reverses the behavior of $f$.
- This one-to-one property guarantees
**$f^{-1}$ is a function**, meaning $f^{-1}$ must pass the vertical line test. - One-to-one functions are the only functions which possess an inverse.
- $f^{-1}$ is not equivalent to $\frac{1}{f}$.

If $f(1) = 5, f(3) = 7, f(8) = 10$, find $f^{-1}(5), f^{-1}(7), f^{-1}(10)$.

$f^{-1}$ must reverse the effect of $f$. Thus \[f^{-1}(5) = 1 \qquad f^{-1}(7) = 3 \qquad f^{-1}(10) = 8\]

Inverse Function Property: Let $f$ be a one-to-one function with domain $A$ and range $B$. The inverse function $f^{-1}$ satisfies the following properties
\begin{align}
f^{-1}(f(x))=x \qquad &\text{ for every } x \text{ in } A\\
f(f^{-1}(x))=x \qquad &\text{ for every } x \text{ in } B
\end{align}

In other words, the composition $(f\circ f^{-1})(x) = x$ and vice versa. This technique is handy to determine whether two functions $f(x)$ and $g(x)$ are inverses of each other.

Show $f(x) = x^3$ and $g(x) = x^{\frac{1}{3}}$ are inverses of each other.

If they are inverses, they must satisfy the inverse function property. Indeed:
\begin{align}
f(g(x)) &= f(x^{\frac{1}{3}}) = (x^{\frac{1}{3}})^3 = x\\
g(f(x)) &= g(x^{3}) = (x^{3})^\frac{1}{3} = x
\end{align}
Finally, notice that both $f(x)$ and $g(x)$ have a domain of $\mathbb{R}$.

Given a function $f$, here are the steps to find the inverse $f^{-1}$:

- Check $f$ is an one-to-one function. If so, proceed. If not, $f$ does not have an inverse.
- Write $y = f(x)$. If there is a domain restriction, convert the domain restriction into a range restriction.
- Solve this equation for $x$ in terms of $y$.
- Interchange $x$ and $y$, including the range restriction into a domain restriction if there was one. The resulting equation is $y = f^{-1}(x)$.

Find the inverse of $f(x) = 3x - 2$.

- $f(x)$ is clearly one-to-one; it is a line sloping from the third quadrant to the first.
- Write $y = 3x - 2$.
- Solve for (isolate) $x$: \begin{align} y &= 3x - 2 \\ y + 2 &= 3x \\ x &= \frac{y+2}{3} \end{align}
- Interchange $x$ and $y$: \[y = \frac{x+2}{3}\]Thus $f^{-1}(x) = \frac{x+2}{3}$.

Find the inverse of $f(x) = \frac{x^5-3}{2}$.

- $f(x)$ is one-to-one by the horizontal line test.
- Write $y = \frac{x^5-3}{2}$.
- Solve for $x$: \begin{align} y &= \frac{x^5-3}{2} \\ 2y &= x^5 - 3 \\ 2y + 3 &= x^5\\ x &= \sqrt[5]{2y+3} \end{align}
- Interchange $x$ and $y$: \[y = \sqrt[5]{2x+3}\]Thus $f^{-1}(x) = \sqrt[5]{2x+3}$.

Find the inverse of $f(x) = \frac{2x + 3}{x - 1}$.

- $f(x)$ is one-to-one by the horizontal line test.
- Write $y = \frac{2x + 3}{x - 1}$.
- Solve for $x$: \begin{align} y &= \frac{2x + 3}{x - 1} \\ y(x-1) &= 2x+3 \\ xy - y &= 2x + 3 \\ xy - 2x &= y + 3 \\ x(y - 2) &= y + 3\\ x &= \frac{y+3}{y-2} \end{align}
- Interchange $x$ and $y$: \[y = \frac{x+3}{x-2}\]Thus $f^{-1}(x) = \frac{x+3}{x-2}$.

Find the inverse of $f(x) = x^2, x \geq 3$.

In class.

Find the inverse of $f(x) = \frac{1}{3}\sqrt{-\left(\frac{1}{2}x + 3\right)}, x \geq -24$.

In class.