A function $y=f(x)$ is said to be one-to-one if satisfies the property
$$f(x_1)=f(x_2) \Longrightarrow x_1=x_2$$
or equivalently
$$x_1\ne x_2 \Longrightarrow f(x_1)\ne f(x_2)$$
for all $x_1,x_2$ in the domain. In plain English what this says is no two numbers in the domain are corresponded to the same number in the range. Figure 1 is the graph of $f(x)=x^2$. It is not one-to-one.

Figure 1. The graph of y=x^2

For example, $-1\ne 1$ but $f(-1)=1=f(1)$.

Figure 2. The graph of y=x^3

Figure 2 is the graph of $f(x)=x^3$. It is one-to-one as seen clearly from the graph. But let us pretend that we don’t know the graph but want to prove that it is one-to-one following the definition. Here we go. Suppose that $f(x_1)=f(x_2)$. Then $x_1^3=x_2^3$ or $x_1^3-x_2^3=(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0$. This means $x_1=x_2$ which completes the proof.

Why do we care about one-to-one functions? The reason is that if $y=f(x)$ is one-to-one, it has an inverse function $y=f^{-1}(x)$.
\begin{align*}
x&\stackrel{f}{\longrightarrow} y\\
x&\stackrel{f^{-1}}{\longleftarrow} y
\end{align*}
Given a one-to-one function $y=f(x)$, here is how to find its inverse function $y=f^{-1}(x)$

STEP 1. Swap $x$ and $y$ in $y=f(x)$. The reason we are doing this is that $\mathrm{Dom}(f)=\mathrm{Range}(f^{-1})$ and $\mathrm{Dom}(f^{-1})=\mathrm{Range}(f)$.

STEP 2. Solve the resulting expression $x=f(y)$ for $y$. That is the inverse function $y=f^{-1}(x)$.

Example. Find the inverse function of $f(x)=\frac{2x+3}{x-1}$. (It is a one-to-one function.)

Solution. STEP 1. Let $y=\frac{2x+3}{x-1}$ and swap $x$ and $y$. Then we have
$$x=\frac{2y+3}{y-1}$$

STEP 2. Let us solve $x=\frac{2y+3}{y-1}$ for $y$. First multiply $x=\frac{2y+3}{y-1}$ by $y-1$. Then we have $x(y-1)=2y+3$ or $xy-x=2y+3$. Isolating the terms that contain $y$ in the LHS, we get $xy-2y=x+3$ or $(x-2)y=x+3$. Finally we find $y=\frac{x+3}{x-2}$. This is the inverse function.

$y=f(x)$ and its inverse $y=f^{-1}(x)$ satisfy the following properties.
$$(f\circ f^{-1})(x)=x,\ (f^{-1}\circ f)(x)=x$$
The reason for these properties to hold is clear from the definition of an inverse function. We can check the properties using the above example. I will do $(f\circ f^{-1})(x)=x$ and leave the other for an exercise.
\begin{align*}
(f\circ f^{-1})(x)&=f(f^{-1}(x))\\
&=f\left(\frac{x+3}{x-2}\right)\\
&=\frac{2\left(\frac{x+3}{x-2}\right)+3}{\left(\frac{x+3}{x-2}\right)-1}\\
&=x
\end{align*}
The graph of $y=f(x)$ and the graph of its inverse $y=f^{-1}(x)$ satisfy a nice symmetry, namely they are symmetric about the line $y=x$. This symmetry helps us obtain the graph of $y=f^{-1}(x)$ when the explicit expression for $f^{-1}(x)$ is not available. You will see such a case later when you study the logarithmic functions. Figure 3 shows the symmetry with $y=x^2$ ($x\geq 0$) and its inverse $y=\sqrt(x)$.

Figure 3. The symmetry between the graphs of y=x^2 (red) and y=sqrt(x) (blue) about y=x (green)

It’s quite interesting that functions can be treated like numbers, namely you can define $+$, $-$, $\times$, and $\div$ on a collection of functions. How do we do this? For instance given two functions $f$ and $g$, we can define a new function $f+g$ by
$$(f+g)(x)=f(x)+g(x)$$
for all $x$ in the domain (for sake of simplicity we assume that $f$ and $g$ both have the same domain. If not, one can take the intersection of the domains of $f$ and $g$, no big deal). In a similar manner, we can also define $f-g$, $fg$, and $\frac{f}{g}$ respectively as
\begin{align*}
(f-g)(x)&=f(x)-g(x)\\
(fg)(x)&=f(x)g(x)\\
\left(\frac{f}{g}\right)(x)&=\frac{f(x)}{g(x)}\ \mbox{provided}\ g(x)\ne 0
\end{align*}

Example. Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$.

(a) Find the functions $f+g$, $f-g$, $fg$ and $\frac{f}{g}$ and their domains.

(b) Find $(f+g)(4)$, $(f-g)(4)$, $(fg)(4)$, $\left(\frac{f}{g}\right)(4)$

Solution. (a) $\mathrm{Dom}(f)=\{x|x\ne 2\}$ and $\mathrm{Dom}(g)=\{x|x\geq 0\}$. So the intersection is $\{x|0\leq x<2\}\cup\{x|x>2\}=[0,2)\cup(2,\infty)$ and this is the domain of $f+g$, $f-g$ and $fg$. For $\frac{f}{g}$ since $g$ is not defined at $x=0$, its domain should be $(0,2)\cup(2,\infty)$.
\begin{align*}
(f+g)(x)&=\frac{1}{x-2}+\sqrt{x}\\
(f-g)(x)&=\frac{1}{x-2}-\sqrt{x}\\
(fg)(x)&=\frac{\sqrt{x}}{x-2}\\
\left(\frac{f}{g}\right)(x)&=\frac{1}{(x-2)\sqrt{x}}
\end{align*}

(b) I will do only $(f+g)(4)$. One way to evaluate $(f+g)(4)$ is to use $(f+g)(x)$ we obtained in part (a) i.e. $(f+g)(4)=\frac{1}{4-2}+\sqrt{4}=\frac{5}{2}$. Another way is evaluating $f(4)$ and $g(4)$ first which are $f(4)=\frac{1}{2}$ and $g(4)=2$. Then $(f+g)(4)=f(4)+g(4)=\frac{5}{2}$.

Composite Functions

Given two functions $f$ and $g$, if the range of $f$ is a subset of the domain of $g$, then we can combine the two functions to create a new function which we will denote by $g\circ f$.
$$x\stackrel{f}{\longmapsto} f(x)\stackrel{g}{\longmapsto} g(f(x))$$
The above diagram hints us that we can define a new function $g\circ f$ by
$$(g\circ f)(x)=g(f(x))$$
We call $g\circ f$ “$f$ followed by $g$.”

Example. Let $f(x)=x^2$ and $g(x)=x-3$.

(a) Find composite functions $f\circ g$ and $g\circ f$ and their domains.

(b) Find $(f\circ g)(5)$ and $(g\circ f)(5)$.

Solution. (a) By definition $(f\circ g)(x)=f(g(x))=f(x-3)=(x-3)^2$. Also by definition $(g\circ f)(x)=g(f(x))=g(x^2)=x^2-3$. From these we can clearly see both their domains are $(-\infty,\infty)$. In general $\mathrm{Dom}(f\circ g)=\mathrm{Dom}(g)$ and $\mathrm{Dom}(g\circ f)=\mathrm{Dom}(f)$.

(b) $(f\circ g)(5)$ can be evaluated using $(f\circ g)(x)$ we obtained in part (a).
$$(f\circ g)(5)=(5-3)^2=4$$
There is another way to do this. If you don’t have to find $(f\circ g)(x)$ but only need to calculate $(f\circ g)(5)$, this may be simpler. First note $(f\circ g)(5)=f(g(5))$. $g(5)=5-3=2$, so $f(g(5))=f(2)=2^2=4$. Similarly we find $(g\circ f)(5)=22$. In general $(f\circ g)(x)\ne (g\circ f)(x)$.

Nonlinear inequalities may seem more complicated and difficult to solve than linear inequalities. However it is not really the case. There is one simple way to solve a nonlinear inequality. It’s called the test point method. I will explain this with an example.

Example. Solve the inequality $x^2\leq 5x-6$.

Solution. The inequality can be rewritten $x^2-5x+6\leq 0$. First we find points at which $x^2-5x+6=0$. Since $x^2-5x+6=(x-2)(x-3)$, $x=2,3$. These two points divide the real line into 3 regions, where $x<2$, where $2<x<3$, and where $x>3$ as shown in Figure 1.

Figure 1. Quadratic Inequality

In each region we pick a test point to see if that test number satisfies the given inequality. If it does, any other number in the same region would satisfy the inequality. If not, any other number in the same region wouldn’t either. While this is pretty cool, you may wonder why this works. One number speaks for the entire numbers in the same region. It’s hard to explain here though but it is due to the continuity of the function $f(x)=x^2-5x+6$. I will leave it at that and will not delve into that any further. You will understand what I said when you learn calculus. In the region $x<2$, I would pick $x=0$ for a test point. But $x=0$ won’t satisfy the inequality as the LHS is $6>0$. Move onto the next region $2<x<3$. I pick $x=2.5=\frac{5}{2}$. Since $\left(\frac{5}{2}\right)^2-5\frac{5}{2}+6=\frac{25-50+24}{4}=-\frac{1}{4}<0$. So this means that $2<x<3$ is a solution of the inequality. In the final region $x>3$ I pick $x=4$. $(4)^2-5(4)+6=2>0$ so no number in this region would satisfy the inequality. Since $x=2$ and $x=3$ also satisfy the inequality, the overall solution is $2\leq x\leq 3$ or $[2,3]$ in interval notation.

The inequalities like one we just did is called quadratic inequalities. For quadratic inequalities we can actually classify solutions depending on inequalities without going through the test point method every time. Let us assume that the quadratic function $f(x)=ax^2+bx+c$ with $a>0$ has two real solutions $\alpha$ and $\beta$ ($\alpha<\beta$). Then the graph of $f(x)$ would look like one in Figure 2. You can find the solution of each of the following quadratic inequalities easily from the graph in Figure 2.

Figure 2. Quadratic Inequality

1. $ax^2+bx+c>0$: $x<\alpha$ or $x>\beta$. In interval notation, $(-\infty,\alpha)\cup(\beta,\infty)$.
2. $ax^2+bx+c\geq 0$: $x\leq\alpha$ or $x\geq\beta$. In interval notation, $(-\infty,\alpha]\cup[\beta,\infty)$.
3. $ax^2+bx+c<0$: $\alpha<x<\beta$. In interval notation, $(\alpha,\beta)$.
4. $ax^2+bx+c\leq 0$: $\alpha\leq x\leq\beta$. In interval notation, $[\alpha,\beta]$.

Let us go over a couple more examples of nonlinear inequalities that are not quadratic inequalities.

Example. Solve $x(x-1)^2(x-3)<0$.

Solution. The method is the same as the first example. We use the test point method. First find $x$ at which $x(x-1)^2(x-3)=0$. They are $x=0, 1, 3$. So there are 4 regions under consideration. $x<0$, $0<x<1$, $1<x<3$, and $x>3$. In the region where $x<0$, the test point $x=-1$ results the sign of the LHS is $+$. So $x<0$ is not a solution. In the region $0<x<1$, the test point $x=\frac{1}{2}$ results the sign of the LHS $-$, so $0<x<1$ is a solution. In the region $1<x<3$, the test point $x=2$ results the sign of the LHS still $-$. This is actually due to $(x-1)^2$. In general if you see an even number of repeated term $x-a$ in your polynomial inequality like the one we have the sign of the polynomial does not change at $x=a$. A little goody to know so you can save time. Let us move onto next and last one. For $x>3$ the test point $x=4$ results the sign of the LHS $+$, so $x>3$ is not a solution. Therefore, the overall solution is $0<x<1$ or $1<x<3$. In interval notation, it is $(0,1)\cup(1,3)$.

Example. Solve $\frac{1+x}{1-x}\geq 1$.

Solution. First rewrite the inequality as $\frac{1+x}{1-x}-1\geq 0$ which simplifies to $\frac{2x}{1-x}\geq 0$. Inequality like this we consider points at which the numerator is 0 and also points at which the denominator is 0. In our case they are $x=0, 1$ and these two points divide the real line into three regions: $x<0$, $0<x<1$, $x>1$. In the region $x<0$, the test point $x=-1$ results the sign of the LHS $-$, so $x<0$ is not a solution. In the region $0<x<1$, the test point $x=\frac{1}{2}$ results the sign of the LHS $+$, so $0<x<1$ is a solution. Finally in the region $x>1$ the test point $x=2$ results the sign of the LHS $-$, so $x>1$ is not a solution. Since $x=0$ also satisfies the inequality, the overall solution is $0\leq x<1$ or $[0,1)$ in interval notation.