- Recall the exponential model for the population of the Earth:
- \(f(t)=(1436.53)\cdot(1.01395)^t\), where \(t\) is the number of years since \(1900\), and \(P=f(t)\) is the population measured in millions of people.
- We saw that it can be used to predict the population at eg. 2020: \(f(t=120)\approx7573.549\).
- Note that \(f(t)\) assigns years \(t\) to population values \(P\).
- Suppose you want to predict when the population will exceed 8 billion.
- This means that you want to reverse the assignment, and assign to population values \(P\) the years \(t\).
- This can be done via the
*inverse function*\(t=f^{-1}(P)\) of \(f(t)\). - The prediction will be \(f^{-1}(P=8000)\approx123.954\), that is the model predicts that population will exceed 8 billion in 2023.

- Recall that a
*function*\(y=f(x)\) assigns precisely one value \(f(x)\) to every \(x\) in its domain.- This means that in order for the inverse \(x=f^{-1}(y)\) to be a function, for each \(y\), there can be at most one \(x\) such that \(f(x)=y\).
*Definition.*A function \(f(x)\) is called*one-to-one*, if it never takes on the same value twice: \(f(x_1)\ne f(x_2)\) whenever \(x_1\ne x_2\).- For example, the function \(f(x)=x^2\) is not one-to-one, because we have \(f(x=-1)=(-1)^2=1=1^2=f(x=1)\).
- On the other hand, the function \(f(x)=x^3\) is one-to-one.
- In the next week, we will learn about continuous functions. In particular, we will learn the Intermediate Value Theorem, which will imply that if the domain \(D\) of a continuous function \(f(x)\) is one connected interval, with its endpoints possibly \(\pm\infty\), then it's one-to-one precisely when it's strictly monotonous.
*Horizontal line test.*A function is one-to-one if and only if no horizontal line intersects its graph more than once.- Exercises: 1.5: 10,12

*Definition.*Let \(y=f(x)\) be a function with domain \(D\) and range \(R\). Then its*inverse function*\(x=f^{-1}(y)\) has domain \(R\) and range \(D\), and it is defined by \[\begin{equation}\tag{*} f^{-1}(y)=x\text{ if and only if }y=f(x)\text{ for any $y$ in $R$}. \end{equation}\]*Caution.*Do not confuse the inverse function \(f^{-1}(x)\) with the reciprocal \(f(x)^{-1}=\frac{1}{f(x)}\). Note that in the inverse function, the \({}^{-1}\) is after the function name, and in the reciprocal, it is after the argument.*Cancellation equations.*By substituting the first equation in the definition (*) to the second and vice versa, we get the following. \[\begin{align*} y=f(f^{-1}(y))&\text{ for any $y$ in $R$}\\ f^{-1}(f(x))=x&\text{ for any $x$ in $D$} \end{align*}\]

- Let \(f(x)\) be a one-to-one function with range \(R\).
- To find its inverse function \(f^{-1}(y)\), you need to solve the equation \(y=f(x)\) for \(x\) for every \(y\) in \(R\).
- Afterwards, you can interchange \(x\) and \(y\) to get the inverse in the form \(f^{-1}(x)\), if you want to.
- Exercises: Find the inverses of \(2x-3\), \(x^3+2\), and \(x^2-x\), the latter with domain \(x\ge\frac12\).
- Click here for the graphs of the functions and their inverses.
- Note that the graph of the inverse \(y=f^{-1}(x)\) is obtained by mirroring the graph of the original function \(y=f(x)\) with respect to the line \(y=x\).
- Exercises: 1.5.22,20