2.16 Symmetries of $f(x)$ - More Transformations (HL)

1. Even and Odd Functions

Symmetry of a function can often be tested algebraically by substituting $-x$ into the function. This means we compare $f(-x)$ with $f(x)$ and $-f(x)$.

Even functions: $$f(-x)=f(x).$$ The graph is symmetrical about the $y$-axis. A basic example is $f(x)=x^2$.
Odd functions: $$f(-x)=-f(x).$$ The graph has rotational symmetry of $180^\circ$ about the origin. A basic example is $f(x)=x^3$.
Neither even nor odd: If $f(-x)$ is neither $f(x)$ nor $-f(x)$, then the function has neither of these symmetries.

IMPORTANT DOMAIN CONDITION

A function can only be even or odd if its domain is symmetrical about $0$. This means that whenever $x$ is in the domain, $-x$ must also be in the domain.

For example, a function with domain $x\geq 0$ usually cannot be classified as even or odd, because the negative side of the domain is missing.
Type Algebraic test Graphical symmetry Example
Even $f(-x)=f(x)$ Symmetry about the $y$-axis $f(x)=x^2$
Odd $f(-x)=-f(x)$ Rotational symmetry about the origin $f(x)=x^3$
Neither Neither condition holds No required symmetry $f(x)=x+x^2$

EXAMPLE 1

Algebraically determine whether each function is even, odd, or neither.

(a) $f(x)=x^4+|x|$. $$\begin{aligned} f(-x)&=(-x)^4+|-x|\\ &=x^4+|x|\\ &=f(x). \end{aligned}$$ Therefore, $f$ is even.
(b) $g(x)=x-x^3$. $$\begin{aligned} g(-x)&=(-x)-(-x)^3\\ &=-x+x^3\\ &=-(x-x^3)\\ &=-g(x). \end{aligned}$$ Therefore, $g$ is odd.
(c) $h(x)=x+x^2$. $$\begin{aligned} h(-x)&=(-x)+(-x)^2\\ &=-x+x^2. \end{aligned}$$ This is neither $h(x)=x+x^2$ nor $-h(x)=-x-x^2$. Therefore, $h$ is neither even nor odd.

GRAPHICAL COMPARISON: EVEN AND ODD SYMMETRY

Even: $y=x^2$
Odd: $y=x^3$

2. Absolute Value Transformations

Applying the modulus symbol changes a graph in different ways depending on whether the modulus is placed around the output or around the input.

The transformation $|f(x)|$: this changes the outputs. All parts of the graph below the $x$-axis are reflected upward across the $x$-axis. $$|f(x)|= \begin{cases} f(x), & f(x)\geq 0,\\ -f(x), & f(x)\lt 0. \end{cases}$$
The transformation $f(|x|)$: this changes the inputs. The graph for $x\geq 0$ is kept, and then reflected into the region $x\lt 0$ across the $y$-axis. $$f(|x|)= \begin{cases} f(x), & x\geq 0,\\ f(-x), & x\lt 0. \end{cases}$$
Transformation What changes? Graphical effect
$|f(x)|$ The $y$-values Negative parts reflect upward across the $x$-axis
$f(|x|)$ The $x$-values Right half of the graph is mirrored across the $y$-axis

EXAMPLE 2

Sketch $y=f(x)$ and $y=|f(x)|$ for $$f(x)=x^2-4.$$

Solution:

The original function is $$f(x)=x^2-4.$$ It is below the $x$-axis when $$\begin{aligned} x^2-4&\lt 0\\ x^2&\lt 4\\ -2&\lt x\lt 2. \end{aligned}$$
Therefore, for $-2\lt x\lt 2$, the graph is reflected upward. The transformed function is $$|f(x)|=|x^2-4|= \begin{cases} x^2-4, & x\leq -2\text{ or }x\geq 2,\\ 4-x^2, & -2\lt x\lt 2. \end{cases}$$

EXAMPLE 3

Sketch $y=f(x)$ and $y=f(|x|)$ for $$f(x)=x+1.$$

Solution:

The transformed function is $$f(|x|)=|x|+1.$$ Hence $$f(|x|)= \begin{cases} x+1, & x\geq 0,\\ -x+1, & x\lt 0. \end{cases}$$
The right-hand part of $y=x+1$ is kept, and then reflected across the $y$-axis.

3. The Reciprocal Function $\dfrac{1}{f(x)}$

The reciprocal transformation changes the output values of a function by replacing each value $y$ with $\dfrac{1}{y}$. Thus, if $$g(x)=\dfrac{1}{f(x)},$$ then $g(x)$ is undefined wherever $f(x)=0$.

To sketch the reciprocal graph $g(x)=\dfrac{1}{f(x)}$, use the following rules:

  1. Roots of $f(x)$ become vertical asymptotes of $\dfrac{1}{f(x)}$.
  2. Vertical asymptotes of $f(x)$ may become roots of $\dfrac{1}{f(x)}$.
  3. If $f(x)$ has horizontal asymptote $y=a$, where $a\neq 0$, then $\dfrac{1}{f(x)}$ has horizontal asymptote $y=\dfrac{1}{a}$.
  4. Points where $f(x)=1$ or $f(x)=-1$ remain unchanged because $\dfrac{1}{1}=1$ and $\dfrac{1}{-1}=-1$.
  5. Positive parts of $f(x)$ remain positive, and negative parts remain negative.
  6. Large values of $|f(x)|$ become small values of $\left|\dfrac{1}{f(x)}\right|$.
Feature of $f(x)$ Feature of $\dfrac{1}{f(x)}$
$x$-intercept of $f(x)$ Vertical asymptote
Vertical asymptote of $f(x)$ Possible $x$-intercept
Horizontal asymptote $y=a$ Horizontal asymptote $y=\dfrac{1}{a}$, if $a\neq 0$
Point with $y=1$ or $y=-1$ Same point remains fixed

EXAMPLE 4

Sketch the reciprocal graph of $$f(x)=x^2-4.$$ That is, sketch $$g(x)=\dfrac{1}{x^2-4}.$$

Solution:

The zeros of $f(x)$ are $$\begin{aligned} x^2-4&=0\\ (x-2)(x+2)&=0\\ x&=2\quad\text{or}\quad x=-2. \end{aligned}$$ Therefore, the reciprocal graph has vertical asymptotes $$x=-2\qquad\text{and}\qquad x=2.$$
As $x\to\pm\infty$, $x^2-4\to\infty$, so $$\dfrac{1}{x^2-4}\to 0.$$ Therefore, the horizontal asymptote is $$y=0.$$
The sign is positive when $x\lt -2$ or $x\gt 2$, and negative when $-2\lt x\lt 2$.

4. The Squared Transformation $[f(x)]^2$

The squared transformation replaces every output $y=f(x)$ by $y=[f(x)]^2$. This makes all output values non-negative.

  • Points where $f(x)=0$ remain at $y=0$.
  • Points where $f(x)=1$ remain at $y=1$.
  • Points where $f(x)=-1$ move to $y=1$.
  • If $|f(x)|\gt 1$, then $[f(x)]^2$ is larger than $|f(x)|$.
  • If $0\lt |f(x)|\lt 1$, then $[f(x)]^2$ is smaller and closer to $0$.
  • All negative parts of $f(x)$ move to the top half-plane after squaring.
Original value $y=f(x)$ New value $[f(x)]^2$ Effect
$0$ $0$ Unchanged
$1$ $1$ Unchanged
$-1$ $1$ Reflected upward
$0\lt y\lt 1$ $0\lt y^2\lt y$ Moves closer to $0$
$y\gt 1$ $y^2\gt y$ Stretches upward
$y\lt 0$ $y^2\gt 0$ Moves to upper half-plane

EXAMPLE 5

Sketch $y=f(x)$ and $y=[f(x)]^2$ for $$f(x)=x-1.$$

Solution:

Since $f(x)=x-1$, the squared transformation is $$[f(x)]^2=(x-1)^2.$$
The original graph is a straight line. After squaring, the graph becomes a parabola with vertex at $(1,0)$.
Useful points are: $$\begin{aligned} f(0)&=-1, & [f(0)]^2&=1,\\ f(1)&=0, & [f(1)]^2&=0,\\ f(2)&=1, & [f(2)]^2&=1. \end{aligned}$$