Functions of single variable

Functions For each variable $x$ in $D \subset {\mathbb{R}}$, there is exactly one $y$ so that the ordered pair $(x,y)$ is contained in the subset defining rule $f$. This rule is called function and denoted by $y = f(x)$.

NOTE A variable in $D$ is called an independent variable, The value $y$ determined by $x$ is called the dependent variable. If $y$ is a function of $x$,

$${\rm Dom}(f) = \{x \in {\mathbb{R}} : f(x) \in {\mathbb{R}}\}$$

is called the domain of $f$ and

$${\rm Range}(f) = \{y \in {\mathbb{R}}: y = f(x), x \in {\rm Dom}(f)\}$$

is called the range of $f$.

Figure 1.1: Not a function

Domain of $f$ The domain of $f$ is the set of variables of $x$ for which $f(x)$ is also a real.

Example 1..1   Find the domain of the function

$$g(x) = \frac{4x^2 - 3x^3}{6x^2 + 3x}$$

For the denominator is 0, the function is not defined. So, we look for the variables $x$ for which the denominators are not 0.

SOLUTION To find the domain of $g(x)$, it is enough to find the set of variables $x$ so that $${\frac{4x^2 - 3x^3}{6x^2 + 3x}}$$ is also real. Note that for $6x^2 + 3x \neq 0$, $g(x)$ is real. $6x^2 + 3x = 3x(2x + 1) \neq 0$ imples that $x \neq 0, -\frac{1}{2}$. Thus,

$${\rm Dom}(g) = \{x \in {\mathbb{R}} : x \neq 0, -\frac{1}{2}\}$$

Figure 1.2: Not a function

Using the intervals' notation, we have

$$(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, 0) \cup (0, \infty)$$

$x \neq 0, -\frac{1}{2}$ means that $x$ can be in any one of the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 0)$, $(0, \infty)$. Thus we use $cup$.

If the number inside the radical is negative, it becomes a complex number.

Figure 1.3: Exercise1-1-1

Exercise 1..1   Find the domain of the function
$${1. \frac{1}{\sqrt{1-x}}}$$ $${2. \sqrt{\frac{4x^2 - 3x^3}{6x^2 + 3x}}}$$

SOLUTION 1. Note that $\frac{1}{\sqrt{1-x}}$ is real whenever the denominator is not 0 and $1-x \geq 0$. With these conditions, we have $1 - x > 0$. Rewriting to get $x < 1$. Using the interval, $(-\infty, 1)$ $\blacksquare$

2. $\sqrt{\frac{4x^2 - 3x^3}{6x^2 + 3x}}$ is real whenver the denominator is not 0 and the inside the radical has to be non-negative. With these conditions,

$$\frac{4x^2 - 3x^3}{6x^2 + 3x} \geq 0$$   and$$6x^2 + 3x \neq 0.$$

To get rid of the denominator, we multiply the both sides by $(6x^2 + 3x)^2$

$$((6x^2 + 3x)^2 \frac{4x^2 - 3x^3}{6x^2 + 3x} = (6x^2 + 3x)(4x^2 - 3x^3) \geq 0.$$

Simplifying to get

$$3x^3(2x+1)(4-3x) \geq 0.$$

Now we solve the equation instead of the inequality, Note that by the condition on the fraction which claims the denominator can not be 0. Then for $x = 0$ and $x = -\frac{1}{2}$, we have the denominators 0. Thus these values do not satisfy the inequality. We put circle on the number line. On the other hand, $x = \frac{4}{3}$ satisfies the inequality, we put dot on the number line to indicate this number is included. Now we check the sign $3x^3(2x+1)(4-3x)$.

Rational Inequality Multiplying both sides of the equation by the square of the denominator, we can get rid of the denominator without changing the inequality sign.

$$\begin{array}{llllllll} \stackrel{+}{\overline{\hspace{1cm}}... ....6cm} \frac{4}{3} &\hspace{-0.6cm} \hspace{1cm}\\ \end{array}$$

Solving inequality To solve $3x^3(2x+1)(4-3x) \geq 0$, we solve $3x^3(2x+1)(4-3x) = 0$. Then we have $x = 0$ , $x = -\frac{1}{2}$, and $x = \frac{4}{3}$

Using the interval, we have

$$(-\infty,-\frac{1}{2}) \cup (0, \frac{4}{3})\ensuremath{ \blacksquare}$$

Graph For a function $f(x)$, the set of points $(x, f(x))$ on the $xy$-plane is called the graph of a function $f(x)$.

graph $y = x^2,x^3+1,x^4-x^2-1$

Figure 1.4: graph

NOTE The graph of the function $f$ is one way to express the rule between two sets. To draw a nice graph, one must know about the critical points, concave up, concave down.

Composite Function For the range of $g(x)$ is in the domain of $f$, the correspondense between $x$ and $f(g(x))$ is called the composite function and denoted by $f \circ g$.

$f \circ g$ $x \underbrace{{\stackrel{g}{\mapsto} g(x) \stackrel{f}{\mapsto}}}_{f\circ g} f(g(x))$

NOTE The range of $g(x)$ has to be in the domain of $f(x)$. Otherwise, $f(g(x))$ can not be defined.

Example 1..2   Let $${f(x) = x^2 - 5}$$, $${g(x) = \frac{1}{x} + 1}$$ . Find $(f \circ g)(x)$ and $(g \circ f)(x)$.

SOLUTION The range of $g(x)$ is $(-\infty,1)\cup (1,\infty)$ and these are in the domain of $f(x)$. Thus,

$$(f \circ g)(x) = f(g(x)) = (g(x))^2 - 5 = (\frac{1}{x} + 1)^2 - 5 \ensuremath{ \blacksquare}$$

Next, since the domain of $g(x)$ is $(-\infty,0)\cup(0,\infty)$, the range of $f(x)$ given by $[-5, \infty)$ is not in the domain of $g(x)$. Then, exclude the value of $f(x)$ which becomes 0, the range of $f(x)$ is in the domain of $g(x)$. So, for $x^2 - 5 \neq 0$, we have

$$(g \circ f)(x) = g(f(x)) = \frac{1}{f(x)} + 1 = \frac{1}{x^2 - 5} + 1$$

$$= \frac{1 + x^2 - 5}{x^2 - 5} = \frac{x^2 -4}{x^2 - 5}\ensuremath{ \blacksquare}$$

$y=g(x)$ Solve $y = \frac{1}{x} + 1$ for $x$, we have $x = \frac{1}{y-1}$. Then $y$ can not take 1.

Figure 1.5: Example1-2

Exercise 1..2   Find the composite function $(f \circ g)(x)$.
1. $${f(x) = \frac{1}{x} - \frac{1}{x+1}, g(x) = \frac{1}{x}}$$
2. $${f(x) = \left\{\begin{array}{cl} 1 - x, & x \leq 0\\ x^2, & x >... ...eft\{\begin{array}{cl} -x, & x < 1\\ 1 + x, & x \geq 1 \end{array} \right.}$$

SOLUTION 1.

$$(f \circ g)(x) = f(g(x)) = \frac{1}{g(x)} - \frac{1}{g(x)+1} = \frac{1}{\frac{1}{x}} - \frac{1}{\frac{1}{x} + 1} = x - \frac{x}{x+1}.$$

$$(g \circ f)(x) = g(f(x)) = \frac{1}{f(x)} = \frac{1}{\frac{1}{x} ... ...1}{x+1}} = \frac{1}{\frac{x+1 - x}{x(x+1)}} = x(x+1)\ensuremath{ \blacksquare}$$

2.

$$f(g(x)) = \left\{\begin{array}{cl} 1 - g(x), & g(x) \leq 0\\ (g(x))^2, & g(x) > 0 \end{array} \right.$$

Figure 1.6: Exercise1-2-1. $f \circ g$

Figure 1.7: Exercise1-2-1. $g \circ g$

Create Composite fct To find $f(g(x))$, the range of $g$ must be in the domain of $f$. Thus, replace $x$ of $f(x)$ by $g(x)$, and check the graph of $g(x)$.

Figure 1.8: Exercise1-2-. $g \circ f$

By looking at the graph of $g(x)$, for $0 \leq x < 1$, we have $g(x) \leq 0$. Also, for $x < 0$ or $x \geq 1$, we have $g(x) > 0$. Thus,

$$f(g(x)) = \left\{\begin{array}{cl} x^2, & x < 0\\ 1 + x, & 0 \leq x \leq 1\\ (1+x)^2, & x > 1 \end{array} \right.$$

Similarly for $g(f(x))$, we obtain

$$g(f(x)) = \left\{\begin{array}{cl} -f(x), & f(x) < 1\\ 1 + f(x), & f(x) \geq 1 \end{array} \right.$$

Now by the graph of $f(x)$, for $0 < x < 1$, we have $f(x) = x^2 < 1$. Also, for $x \leq 0$, we have $f(x) = 1- x \geq1$, and for $x \geq 1$, we have $f(x) = x^2 \geq 1$. Therefore,

$$g(f(x)) = \left\{\begin{array}{cl} 2-x, & x \leq 0\\ -x^2, & 0 < x < 1\\ 1 + x^2, & x \geq 1 \end{array} \right. \ensuremath{ \blacksquare}$$

One-to-one Function For any $x_{1},x_{2} \in {\rm Dom}(f)$,

$$x_{1} \neq x_{2} \Longrightarrow f(x_{1}) \neq f(x_{2})$$

Then $f$ is said to be one-to-one.

Figure 1.9: Graph Test

The graph of $y = f(x)$ is intersected with more than two points with the line $y = 1$. Thus, it is not one-to-one.

The contrapositive of the statement $x_{1} \neq x_{2} \Longrightarrow f(x_{1}) \neq f(x_{2})$ is $f(x_{1}) = f(x_{2}) \Longrightarrow x_{1} = x_{2}$ Thus, once we show that the contrapositive is true whenever the original statement is true, we can use the contrapositive. A statement is an assertion that can be determined to be true or false. We use $p,q$ for statements. The statement $p \longrightarrow q$ becomes false only if $p$ is true and $q$ is false.

Contrapositive Truth Table
$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert} \hlin... ...T & T & T & F\ \hline F & F & T & T & T\ \hline \end{array}$$ The contrapositive of the statement $p \Rightarrow q$ is given as $\bar{q} \Rightarrow \bar{p}$ which is equivalent to $p \Rightarrow q$.

Example 1..3   Find the following functions are on-to-one or not.
$${ 1. f(x) = \frac{x}{\vert x\vert}}$$ $${2. f(x) = \frac{1}{x+2}, -2 < x}$$

SOLUTION 1. For $x_{1} = 1, x_{2} = 2$, we have $f1. = 1 = f2.$. Thus, it is not on-to-one $\blacksquare$
2. Suppose that $f(x_{1}) = f(x_{2})$. Then $\frac{1}{x_{1} + 2} = \frac{1}{x_{2} + 1}$. Multiply $(x_{1} + 2)(x_{2} + 2)$ to the both sides, we have $x_{1} + 2 = x_{2} + 2$. Thus $x_{1} = x_{2}$ $\blacksquare$

Exercise 1..3   Find the following function is one-to-one.

$$y = f(x) = x^3 + 1$$

SOLUTION We show $f(x_{1}) = f(x_{2}) \longrightarrow x_{1} = x_{2}$. $f(x_{1}) = f(x_{2})$ implies that $x_{1}^3 + 1 = x_{2}^3 + 1$ which implies that $x_{1}^3 = x_{2}^3$. Now we have show $x_{1} = x_{2}$ is the only solutioin. To show this, we write $x_{1}^3 - x_{2}^3 = (x_{1} - x_{2})(x_{1}^2 + x_{1}x_{2} + x_{2}^2) = 0$. Then we have $x_{1}^2 + x_{1}x_{2} + x_{2}^2 = (x_{1} + \frac{x_{2}}{2})^2 + \frac{3x_{2}^2}{4}$. This is sums of squres. Thus they are never 0 except $x_{1} = x_{2} = 0$. This shows that $x_{1} = x_{2}$ is the only solution. $\blacksquare$

one-to-one To find the given function is one-to-one, it is enough to show $f(x_{1}) = f(x_{2}) \longrightarrow x_{1} = x_{2}$. To show it is not one-to-one, it is enough to give one counter example.

Inverse Function For a function $f$ is one-to-one, the correspondence $f$ between each $y \in {\mathbb{R}}(f)$ and unique $x$ such that $x = g(y)$ is called the inverse function of $f$ and denoted by $f^{-1}$.

NOTE The inverse function of $y = f(x)$ is $x = g(y)$ and satisfies $y = f(x) = f(g(y))$. Thus we can write

$$y = f^{-1}(x) \Leftrightarrow x = f(y).$$

From this, to find the inverse function of $y = f(x)$, we can simply change $x$ and $y$ and solve for $y$. This way we can find the inverse function of $y = f(x)$.

Symmetric The inverse function $y = f^{-1}(x)$ and the function $y = f(x)$ is symmetric with respect to the line $y = x$.

Example 1..4   Show the following function is one-to-one. Then find the inverse function.

$$y = f(x) = \frac{3x + 1}{x - 2}, 2 < x < \infty$$

SOLUTION Suppose that $f(x_{1}) = f(x_{2})$. Then $\frac{3x_{1} + 1}{x_{1} - 2} = \frac{3x_{2} + 1}{x_{2} - 2}$. Now multiply both sides by $(x_{1} - 2)(x_{2} -2)$ and simplify the equation to get

$$(3x_{1} + 1)(x_{2} - 2) = (3x_{2} + 1)(x_{1} - 2).$$

How to find the inverse We put $f(y) = x$ and solve for $y$.

$$3x_{1}x_{2} - 6x_{1} + x_{2} - 2 = 3x_{1}x_{2} + x_{1} - 6x_{2} - 2.$$

Simplifying the equation to obtain $-7x_{1} +7x_{2} = 0$. Thus $x_{1} = x_{2}$.

Figure 1.10: Expansion

Next we find the inverse $f^{-1}$. Using $f(f^{-1}(x)) = x$ to obtain

$$f(f^{-1}(x)) = f(y) = \frac{3y + 1}{y - 2} = x$$

Solve this for $y$, we get $3y + 1 = xy - 2x$. Then $xy-3y = 2x+1$ and $${y = f^{-1}(x) = \frac{2x + 1}{x - 3}}$$ $\blacksquare$How to find the inverse
1. Replace $x$ by $y$.
2. Solve for $y$.

Exercise 1..4   Determine the following function is one-to-one or not. If so, find the inverse of the function.
$${1. f(x) = \frac{1}{1-x} - 1}$$ $${2. f(x) = \frac{1}{(x+1)^{2/3}}}$$

SOLUTION 1. Suppose that $f(x_{1}) = f(x_{2})$. Then $\frac{1}{1-x_{1}} - 1 = \frac{1}{1-x_{2}} - 1$. Clearing denominators, we have $1 - x_{1} = 1 - x_{2}$ which implies that $x_{1} = x_{2}$. Thus one-to-one. We next find the inverse function. Replace $x$ by $y$, we get $f(y) = \frac{1}{1-y} - 1 = x$ which implies that $\frac{1}{1-y} = x+1$. Now take the reciprocal of both sides, we have $1 - y = \frac{1}{x+1}$ which implies that $y = 1 - \frac{1}{x+1}\ensuremath{ \blacksquare}$
2. $f(0) = 1, f(-2) = \frac{1}{(-1)^{2/3}} = \frac{1}{(-1)^2} = 1$. Thus this is not one-to-one $\blacksquare$

Figure 1.11: Exercise1-4-2's graph

$(-1)^{2/3}$ $$\begin{array}{l} (-1)^{2/3}\\ = ((-1)^{1/3})^2 \\ = (-1)^2 \\ = 1\end{array}$$

ExercisesA

1.

Find the value fo $f(x)$ when the value of $x$ is 1

(a) $${f(x) = \vert 2-x\vert}$$ (b) $${f(x) = 4 + 10x - x^{2}}$$ (c) $${f(x) = 1 + \cos{(x-1)}}$$

2.

Find the domain and range of the following functions

(a) $${f(x) = x^{2} - 1}$$ (b) $${g(x) = \sqrt{1-x}}$$ (c) $${h(x) = \vert\sin{x}\vert}$$

3.

Using the graphs of $y = \frac{1}{x}$ and $y = \sqrt{x}$, draw the graph of the following functions (a) $${y = \frac{1}{x} + 1}$$ (b) $${y = \sqrt{1 - x}}$$

4.

Find the composite functions: $(f \circ g)(x)$ and $g(f(x))$

(a) $${f(x) = 2x+5, g(x) = x^{2}}$$(b) $${f(x) = \frac{1}{x}, g(x) = \frac{1}{x}}$$

(c) $${f(x) = \frac{1}{x} - \frac{1}{x+1}, g(x) = \frac{1}{x^{2}}}$$

5.

Determine the following functions are one-to-one. If so, find the inverse.

(a) $${f(x) = 7x - 4}$$(b) $${f(x) = (x+1)^{3} + 2}$$(c) $${f(x) = \frac{x}{\vert x\vert}}$$

6

A function $f(x)$ is called even function provided $f(-x) = f(x)$ in $x \in D(f)$. On the other hand if $f(-x) = -f(x)$, then the function is called odd function. Determine the following functions are even or odd functions.

(a) $${f(x) = x^{3}}$$(b) $${f(x) = x(x^{2} + 1)}$$

ExercisesB

1.

Determine the following rule gives rise a single-valued function or not

(a) $${y^{2} = x, x > 0}$$ (b) $${y^{3} = x^2}$$

2.

Find the domain of the following functions. Then draw the graph of $f(x)$

(a) $${f(x) = \sqrt{4 - x^2}}$$ (b) $${h(x) = \sqrt{\frac{4x^2 - 3x^3}{6x^2 + 3x}}}$$

3.

Find the composite functions: $(f \circ g)(x)$ and $g(f(x))$

(a) $${f(x) = 2x - 1, g(x) = x^2 + 1}$$

(b) $${f(x) = \left\{\begin{array}{cl} 1 - x, & x \leq 0\\ x^2, & x >... ...eft\{\begin{array}{cl} -x, & x < 1\\ 1 + x, & x \geq 1 \end{array} \right.}$$

4.

Find the inverse of the following functions.

(a) $${f(x) = \frac{1}{x+2}, -2 < x}$$ (b) $${f(x) = x^2 + 4x - 2}$$

5.

Determine the following functions are even or odd functions

(a) $${f(x) = \frac{x^{2}}{1 - \vert x\vert}}$$ (b) $${f(x) = \sin{x}}$$

6.

Determine the following function is even or odd

(a) A product of even function and odd function