2.14 Odd and even functions; inverse with domain restrictionIB Maths: Analysis and Approaches HL: Revision notes

A function is even if $f(-x)=f(x)$ for every $x$ in its domain, and odd if $f(-x)=-f(x)$ for every $x$. The domain must be symmetric about 0 for either to make sense.

• Even: $x^{2}$, $x^{4}+1$, $|x|$, $\cos x$. The graph has reflective symmetry in the $y$-axis.
• Odd: $x$, $x^{3}-4x$, $\frac{1}{x}$, $\sin x$, $\tan x$. The graph has rotational symmetry of order 2 about the origin.
• Most functions are neither, for example $x^{2}+x$.

To test, substitute $-x$ and simplify, then compare with $f(x)$ and with $-f(x)$. To prove a function is not even or odd, one numerical counterexample is enough: for $h(x)=x^{2}+x$, $h(-1)=0$ but $h(1)=2$ and $-h(1)=-2$.

Think of odd and even like signs in multiplication:

• odd $\times$ odd $=$ even, e.g. $x\sin x$
• even $\times$ even $=$ even
• odd $\times$ even $=$ odd, e.g. $x\cos x$
• odd $+$ odd $=$ odd; even $+$ even $=$ even; odd $+$ even is usually neither.

Periodic functions can be odd or even too: $\sin$ and $\tan$ are odd, $\cos$ is even, and $\sin(2x)$, $\cos(3x)$ keep the parity of the original. So $k(x)=\sin2x+x\cos x$ is odd, and $k\left(-\frac{\pi}{4}\right)=-k\left(\frac{\pi}{4}\right)$.

A function has an inverse only if it is one-to-one: each output comes from exactly one input. Algebraically, if $f(a)=f(b)$ forces $a=b$, then $f$ is one-to-one. A function with a turning point inside its domain is many-to-one, so it has no inverse there.

To find $f^{-1}(x)$:

1. Write $y=f(x)$.
2. Rearrange to make $x$ the subject (or swap $x$ and $y$ first).
3. Write the result as $f^{-1}(x)$.

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$.

If $f$ is many-to-one, restrict its domain to a part where it is one-to-one. For a quadratic, complete the square and cut at the vertex.

Example: $h(t)=20-(t-3)^{2}$ has its maximum at $t=3$. On $t\ge3$ it is decreasing, so it has an inverse. From $(t-3)^{2}=20-x$ we get $t=3\pm\sqrt{20-x}$, and we choose the positive root because $t\ge3$: $h^{-1}(x)=3+\sqrt{20-x}$, for $0\le x\le20$ if the stone lands when $h=0$.

The largest domain of the form $x\ge k$ that gives an inverse starts at the turning point.

A function is self-inverse if $f^{-1}(x)=f(x)$, equivalently $f(f(x))=x$ for all $x$ in the domain. Its graph is symmetric in the line $y=x$.

Examples: $f(x)=\frac{1}{x}$, $f(x)=a-x$, and any $f(x)=\dfrac{ax+b}{cx-a}$ with $c\neq0$ and $a^{2}+bc\neq0$.

For $g(x)=\dfrac{3x+5}{x-3}$: $g(g(x))=\dfrac{3(3x+5)+5(x-3)}{(3x+5)-3(x-3)}=\dfrac{14x}{14}=x$, so $g$ is self-inverse. Its domain and range are both $\mathbb{R}\setminus\{3\}$.