Functions

The domain is the set of inputs the function is defined for; the range is the set of outputs it produces over that domain. P1 states a domain in the question and asks for the range, and the reliable method for a quadratic is to complete the square:

The square is never negative, so the least value is the constant — 2, reached at $x=3$. If the domain excludes the vertex the answer changes: on $x\ge 4$ the function is increasing throughout, so the range starts at $\mathrm{f}(4)=3$ and the answer is $\mathrm{f}(x)\ge 3$. Always check whether the given domain contains the turning point before quoting the vertex.

Notation matters for the mark: the range is written in terms of $\mathrm{f}(x)$ or $y$, not $x$. Writing $x\ge 2$ for a range is marked wrong even when 2 is right.

$\mathrm{g}\mathrm{f}(x)$ means “do $\mathrm{f}$ first, then $\mathrm{g}$” — the function nearer $x$ acts first. With $\mathrm{f}(x)=2x+1$ and $\mathrm{g}(x)=x^2$:

They are different, and questions exploit that. A composite exists only if the range of the inner function lies inside the domain of the outer one — so if $\mathrm{g}$ is defined only for $x\ge 0$, then $\mathrm{g}\mathrm{f}$ needs $\mathrm{f}(x)\ge 0$, which restricts the composite’s own domain. “Solve $\mathrm{g}\mathrm{f}(x)=25$” is the standard follow-up: form the composite, set it equal, solve, then reject any root outside the domain.

The inverse undoes the function: ${\mathrm{f}}^{-1}\mathrm{f}(x)=x$. It exists only when the function is one-to-one — each output coming from exactly one input. A quadratic on its natural domain is two-to-one, which is why P1 questions restrict it (to $x\ge 3$, say, one side of the vertex) before asking for an inverse.

To find it: write $y=\mathrm{f}(x)$, make $x$ the subject, then swap the letters.

The $+$ is a decision, not a default: the domain $x\ge 3$ means the outputs of the inverse must be $\ge 3$, so the positive root is the right one. Had the restriction been $x\le 3$, the answer would be $3-\sqrt{x-2}$.

Two facts that questions test directly: the domain of ${\mathrm{f}}^{-1}$ is the range of $\mathrm{f}$ (here $x\ge 2$), and the graph of $y={\mathrm{f}}^{-1}(x)$ is the reflection of $y=\mathrm{f}(x)$ in the line $y=x$. That reflection is also why solving $\mathrm{f}(x)={\mathrm{f}}^{-1}(x)$ can be done by solving $\mathrm{f}(x)=x$ instead.

The largest single sub-skill in this section. Two families, and the rule is that changes inside the bracket affect $x$ and do the opposite of what they look like:

Combined transformations are where the marks are, and order matters. To describe the move from $y=x^2$ to $y=3(x-2{)}^2+1$: translate 2 right, stretch by factor 3 parallel to the $y$-axis, then translate 1 up. Describing a translation needs the word “translation” and a vector; describing a stretch needs the factor and the direction. “Moved up 1” does not score.

1. 1.Range written in terms of x

   A range describes outputs: $\mathrm{f}(x)\ge 3$ or $y\ge 3$. $x\ge 3$ is a domain, and loses the mark.

2. 2.fg and gf the wrong way round

   The function next to the $x$ goes first. $\mathrm{g}\mathrm{f}(x)=\mathrm{g}(\mathrm{f}(x))$ — substitute $\mathrm{f}$ into $\mathrm{g}$.

3. 3.Taking ± on an inverse

   An inverse is a function, so it takes one sign. Choose it from the restricted domain: outputs of ${\mathrm{f}}^{-1}$ must lie in the domain of $\mathrm{f}$.

4. 4.Not stating the inverse's domain

   It is the range of the original function, and it carries its own mark. Part (i) usually hands it to you.

5. 5.f(x + 2) translated right

   Inside the bracket, the effect is reversed: $\mathrm{f}(x+2)$ moves the graph 2 to the left. Same for stretches: $\mathrm{f}(2x)$ is a stretch of factor $\frac{1}{2}$.

6. 6.Keeping a root outside the domain

   After solving a composite equation, test each root against the domain of the inner and outer functions and reject what does not fit.