What if you wanted to go backward? Meaning, what if you wanted to plug in temperature to get the time?
You can, but the relationship needs to be a function. Each temperature can only be sent to one single time point!
To guarantee this, the original time vs. temperature needs to be one-to-one.
One-to-One Functions
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 $f(x_1) \neq f(x_2)$.
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.
Determine which of the following functions are one-to-one: \[f(x) = \sqrt{x}\qquad g(x) = x^2 \qquad h(x) = \frac{1}{x}\] 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.
The Inverse of a Function
Before defining the inverse, we need to define the set of all possible outputs of a function.
codomain
The codomain of a function $f$ is the set of all possible outputs when $f$ is evaluated on the domain.
Sometimes, the codomain is called range. But we will not.
Find the codomain of $f(x) = x^2$.
The domain is $\mathbb{R}$. Squaring any real number results in a number $0$ or greater. Thus the codomain is $[0, \infty)$.
inverse function, $f^{-1}$
Let $f$ be a one-to-one function with domain $A$, codomain $B$. Then the inverse function, denoted $f^{-1}$, has domain $B$ and codomain $A$ and is defined by \[f^{-1}(y) = x \text{ if and only if } f(x) = y\]
Note:
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 $f$.
The 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 $\frac{1}{f}$. Don't confuse it with a negative exponent!
$f^{-1}$ must reverse the effect of $f$. Thus \[f^{-1}(5) = 1 \qquad f^{-1}(7) = 3 \qquad f^{-1}(10) = 8\]
The next property tells you what happens when you take the composition of $f$ and its inverse $f^{-1}$.
Inverse Function Property
Let $f$ be a one-to-one function with domain $A$ and codomain $B$. 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}
Use this technique to determine whether two functions $f(x)$ and $g(x)$ are inverses of each other.
Determine whether $f(x) = x^3$ 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 $\mathbb{R}$.
Determine whether $f(x) = \dfrac{x - 2}{5}$ and $g(x) = -5x + 2$ are inverses of each other.
Finding the Inverse
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 codomain restriction.
Solve this equation for $x$ in terms of $y$. Meaning, isolate $x$.
Interchange $x$ and $y$, including the codomain 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 $f(x) = \frac{1}{3}\sqrt{-\left(\frac{1}{2}x + 3\right)}, x \geq -24$.