Some functions are built so that one half of the graph is a perfect copy of the other half, and spotting that saves you real work.
If you know a curve is symmetric, you only ever have to study half of it, then mirror or spin what you found to get the rest.
Two symmetries show up again and again: a graph that reflects onto itself in the $y$-axis, and a graph that maps onto itself under a half-turn about the origin.
These are exactly the even and odd functions, and this section gives you a one-line algebra test for each so you never have to guess from a sketch.
The same topic then asks the reverse question: when can you run a function backwards to get an inverse, and what do you do when the function is not one-to-one?
The neat payoff at the end is the self-inverse function, a curve so symmetric that running it forwards and running it backwards are the same thing.
Analogy
Think of a butterfly: the left wing is the mirror image of the right, so once you have drawn one wing you already know the other.
An even function is that butterfly, mirrored across the $y$-axis.
An odd function is different: it is like a pinwheel, where turning the whole picture half a turn about the centre lands it exactly back on itself.
Even Functions Mirror Across The y-Axis
A function is even if replacing $x$ with $-x$ leaves the output unchanged, for every $x$ in its domain: $$ f(-x) = f(x) $$
Read this geometrically: the point $(x, f(x))$ and the point $(-x, f(x))$ sit at the same height, one on each side of the $y$-axis.
So the two halves of the graph are reflections of each other in the $y$-axis, which is the visual signature of an even function.
Example
The simplest example is $f(x) = x^2$, because squaring destroys the sign: $$ f(-x) = (-x)^2 = x^2 = f(x) $$
The same works for any even power, for $f(x) = |x|$, for $f(x) = \cos x$, and for any constant function, all of which are symmetric about the $y$-axis.
Hint
The word even is a memory aid: functions built only from even powers of $x$ (including $x^0$, a constant) are even.
As soon as a single odd power sneaks in, that symmetry usually breaks.
Testing A Function Is Even
To test any function, compute $f(-x)$ by substituting $-x$ everywhere $x$ appears, then simplify.
If the simplified result is identical to the original $f(x)$, the function is even.
Example question
Show that $f(x) = x^4 - 3x^2 + 5$ is even.
Solution
Substitute $-x$: $$f(-x) = (-x)^4 - 3(-x)^2 + 5$$
Simplify each power: $(-x)^4 = x^4$ and $(-x)^2 = x^2$, so $f(-x) = x^4 - 3x^2 + 5$.
This is exactly $f(x)$, so $f(-x) = f(x)$ and $f$ is even.
Odd Functions Have Half-Turn Symmetry About The Origin
A function is odd if replacing $x$ with $-x$ flips the sign of the output, for every $x$ in its domain: $$ f(-x) = -f(x) $$
Geometrically, the point $(x, f(x))$ has a partner at $(-x, -f(x))$, which is the original point rotated $180^\circ$ about the origin.
So an odd function has rotational symmetry of order 2 about the origin: a half-turn maps the graph exactly onto itself.
Example
The simplest example is $f(x) = x^3$, because cubing keeps the sign: $$ f(-x) = (-x)^3 = -x^3 = -f(x) $$
The same holds for any odd power, for $f(x) = \tfrac{1}{x}$, and for $f(x) = \sin x$, all of which spin onto themselves under a half-turn.
Note
If an odd function is defined at $x = 0$, its graph must pass through the origin.
Setting $x = 0$ in $f(-x) = -f(x)$ gives $f(0) = -f(0)$, which forces $f(0) = 0$.
Testing A Function Is Odd
Compute $f(-x)$ as before, then compare it with $-f(x)$, which is the original with every term's sign reversed.
If $f(-x)$ and $-f(x)$ match, the function is odd.
Compare with $f(x) = 2x^2 - x$: not equal, so $f$ is not even.
Compare with $-f(x) = -2x^2 + x$: not equal, so $f$ is not odd.
Since it matches neither, $f$ is neither odd nor even.
Finding An Inverse Function
The inverse function $f^{-1}$ undoes the effect of a function, from SL 2.2: if $f$ sends $x$ to $y$, then $f^{-1}$ sends $y$ back to $x$.
The defining relationship is that composing a function with its inverse returns the input unchanged: $$ f\left(f^{-1}(x)\right) = x \quad \text{and} \quad f^{-1}\left(f(x)\right) = x $$
On a graph this makes $f^{-1}$ the reflection of $f$ in the line $y = x$, since reflecting swaps the roles of input and output.
To find a formula for the inverse, follow four steps.
Write $y = f(x)$.
Swap $x$ and $y$, which encodes the reflection in $y = x$.
Solve the new equation for $y$.
Rewrite $y$ as $f^{-1}(x)$.
Example question
Find the inverse of $f(x) = x^3 + 1$.
Solution
Write $y = x^3 + 1$.
Swap $x$ and $y$: $x = y^3 + 1$.
Solve for $y$: $y^3 = x - 1$, so $y = \sqrt[3]{x - 1}$.
Therefore $f^{-1}(x) = \sqrt[3]{x - 1}$.
Tip
Check your inverse by composing: $f^{-1}(f(x))$ should collapse back to $x$.
Here $f^{-1}(f(x)) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3} = x$, which confirms the answer.
Domain Restriction Rescues A Lost Inverse
An inverse exists only when the function is one-to-one, from SL 2.5: every output must come from exactly one input.
A many-to-one function fails this, because two different inputs share an output and the reverse mapping cannot decide which one to return.
The classic offender is $f(x) = x^2$, where both $x = 3$ and $x = -3$ give $9$, so there is no single value $\sqrt{9}$ can reverse to.
The fix is domain restriction: cut the domain down to a piece on which the function is one-to-one, and an inverse appears.
Restricting $f(x) = x^2$ to $x \geq 0$ keeps only the right-hand half of the parabola, which passes the one-to-one test.
On that restricted domain the inverse is the positive square root: $$ f(x) = x^2, \; x \geq 0 \quad \Longrightarrow \quad f^{-1}(x) = \sqrt{x} $$
Restricting instead to $x \leq 0$ is equally valid and gives $f^{-1}(x) = -\sqrt{x}$, so the choice of restriction decides which inverse you get.
Common Mistake
The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$, so state these to keep them consistent.
A common error is giving $f^{-1}(x) = \sqrt{x}$ but forgetting that it is only valid because the domain was restricted to $x \geq 0$.
Choose the restricted domain so that its range still covers the values you need; the wrong half can lose the inverse you actually wanted.
Example question
The function $f(x) = (x - 3)^2$ is defined on the restricted domain $x \geq 3$. Find $f^{-1}(x)$ and state its domain.
Swap: $x = (y - 3)^2$, and take the positive root since $y - 3 \geq 0$: $\sqrt{x} = y - 3$.
Solve for $y$: $y = 3 + \sqrt{x}$, so $f^{-1}(x) = 3 + \sqrt{x}$.
The domain of $f^{-1}$ is the range of $f$, namely $x \geq 0$.
Self-Inverse Functions Are Their Own Reflection In y = x
A function is self-inverse when it equals its own inverse, so running it twice returns the start: $$ f = f^{-1}, \quad \text{equivalently} \quad f\left(f(x)\right) = x $$
Because $f^{-1}$ is the reflection of $f$ in the line $y = x$, a self-inverse graph is unchanged by that reflection: it is its own mirror image in $y = x$.
Every self-inverse function is automatically one-to-one, so its inverse is guaranteed to exist.
The cleanest example is the reciprocal function $f(x) = \tfrac{1}{x}$ from SL 2.8, which reflects onto itself in $y = x$. $$ f\left(f(x)\right) = f\!\left(\tfrac{1}{x}\right) = \frac{1}{\,1/x\,} = x $$
Two more families are worth memorising as self-inverse.
Reflections of the form $f(x) = a - x$
Applying it twice gives $f(f(x)) = a - (a - x) = x$, so any such line is self-inverse.
Rational functions of the form $f(x) = \dfrac{ax + b}{cx - a}$
The key structural feature is that the coefficient of $x$ on top, $a$, is the negative of the constant on the bottom, $-a$.
This symmetry is exactly what makes the double application $f(f(x))$ simplify back to $x$.
Note
Self-inverse is a strong condition, so most invertible functions are not self-inverse.
For instance $f(x) = 2x + 3$ has inverse $\tfrac{x - 3}{2}$, which is a different function, so it is not self-inverse.
Example question
Show that $f(x) = \dfrac{2x + 3}{x - 2}$ is self-inverse.
Solution
Compute $f(f(x))$ by substituting $f(x)$ into itself; write $u = \dfrac{2x + 3}{x - 2}$.
