Further Work with Functions

Further work with functions extends the Advanced focus area Working with Functions. It is one of the more heavily examined Year 11 topics in Extension 1 (roughly 5–7 marks a paper), usually as multiple choice about graphs plus a short Section II inequality or inverse function question.

• Graphical relationships: sketch $y=\frac{1}{f(x)}$, $y=|f(x)|$, $y=f(|x|)$, $y=f(x)+g(x)$ and $y=f(x)-g(x)$ from the graph or equation of $f$ (and $g$); find domains and ranges of sums and differences.
• Reciprocal trig graphs: $y=\sec x$, $y=\mathrm{cosec}x$ and $y=\cot x$ in radians and degrees, with asymptotes, period, domain, range and symmetry.
• Inverse functions: one-to-one functions, the horizontal line test, finding $f^{-1}(x)$, restricting a domain, swapping domain and range, and intersections of $f$ and $f^{-1}$.
• Parametric form: convert lines, parabolas and circles between parametric and Cartesian form and sketch them.
• Inequalities: cubics in factored form, rational inequalities with $x$ in the denominator, and absolute value inequalities.

Reciprocal function

To sketch $y=\frac{1}{f(x)}$ from $y=f(x)$, work point by point:

• Where $f(x)=0$, the reciprocal has a vertical asymptote.
• Where $f(x)$ has a vertical asymptote, the reciprocal approaches $0$ (draw an open circle on the axis there, because the point is still not in the domain).
• The sign is the same: above the axis stays above, below stays below.
• Where $f(x)=\pm 1$ the two graphs meet.
• A maximum of $f$ becomes a minimum of $\frac{1}{f}$ at the same $x$ value, and vice versa (if the value is non-zero).
• If $f(x)\to \pm \infty$ as $x\to \pm \infty$, then $\frac{1}{f(x)}\to 0$, giving a horizontal asymptote $y=0$.

The reciprocal trig functions are the same idea applied to $\sin$, $\cos$ and $\tan$:

$y=\sec x$ and $y=\mathrm{cosec}x$ have period $2\pi$ and range $y\le -1$ or $y\ge 1$; $\sec x$ has asymptotes where $\cos x=0$ ($x=\frac{\pi }{2}+k\pi$) and is even, while $\mathrm{cosec}x$ has asymptotes at $x=k\pi$ and is odd. $y=\cot x$ has period $\pi$, asymptotes at $x=k\pi$, zeros at $x=\frac{\pi }{2}+k\pi$, all real numbers as its range, and it is odd and decreasing on each branch. ($\cot x$ is defined as $\frac{\cos x}{\sin x}$, so it does equal 0 where $\tan x$ is undefined.)

Absolute value graphs

• $y=|f(x)|$: keep the parts of the graph on or above the $x$-axis and reflect the parts below the axis up into it. The result is never below the axis.
• $y=f(|x|)$: keep the part of the graph for $x\ge 0$, delete the left half, then reflect the right half in the $y$-axis. The result is always even.

Do not mix these up. $|f(x)|$ changes the outputs (a vertical change); $f(|x|)$ changes the inputs (a horizontal change).

Sums and differences

To graph $y=f(x)+g(x)$, add the heights of the two graphs at each $x$. Useful checkpoints: where one graph is zero the sum equals the other graph; where the graphs are equal and opposite the sum is zero. The domain of $f+g$ or $f-g$ is the set of $x$ values where both $f$ and $g$ are defined. For example, $\sqrt{x}+\ln (3-x)$ needs $x\ge 0$ and $x<3$, so its domain is $0\le x<3$. Where possible, also state the range of $f+g$ or $f-g$, reading it from your graph (and checking with a graphing application if one is allowed).

A function is one-to-one if every $y$ value in its range comes from exactly one $x$ value. Reflecting in the line $y=x$ swaps the coordinates $(a,b)\to (b,a)$, so the inverse relation is the reflection of the graph in $y=x$. That reflection is a function only if the original passes the horizontal line test (no horizontal line cuts it more than once).

To find $f^{-1}(x)$: write $y=f(x)$, swap $x$ and $y$, then make $y$ the subject. The domain of $f^{-1}$ is the range of $f$ and the range of $f^{-1}$ is the domain of $f$. Formally, $f^{-1}$ is the inverse of $f$ when

on the appropriate domains. If $f$ is not one-to-one, restrict its domain (usually to one side of a turning point) before finding the inverse.

Worked example 1: inverse of a restricted quadratic

Let $f(x)=x^2+4x$ for $x\ge -2$. Find $f^{-1}(x)$, state its domain and range, and find where $y=f(x)$ meets $y=f^{-1}(x)$.

Why it has an inverse. Completing the square, $f(x)=(x+2{)}^2-4$. The vertex is at $(-2,-4)$, and $x\ge -2$ keeps only the right half of the parabola, which is increasing, so $f$ is one-to-one with range $y\ge -4$.

Find the inverse. Swap $x$ and $y$:

Take the positive root because the range of the inverse must be $y\ge -2$ (the domain of $f$). So $f^{-1}(x)=-2+\sqrt{x+4}$ with domain $x\ge -4$ and range $y\ge -2$.

Intersections. For an increasing function, $f$ and $f^{-1}$ can only meet on the line $y=x$, so solve $f(x)=x$:

$x=-3$ is outside the domain $x\ge -2$, so the only intersection is $(0,0)$.

Warning: the shortcut "solve $f(x)=x$" only works when $f$ is increasing. A decreasing function can meet its inverse at points off the line $y=x$.

A curve can be given by two parametric equations $x=x(t)$ and $y=y(t)$, where $t$ is the parameter. To find the Cartesian equation, eliminate $t$:

• For a line or parabola, make $t$ the subject of the simpler equation and substitute.
• For a circle, isolate $\cos \theta$ and $\sin \theta$ and use ${\cos }^2\theta +{\sin }^2\theta =1$.

Standard forms to recognise: the line $x=x_1+at$, $y=y_1+bt$; the parabola $x=2at$, $y=a{t}^2$ (giving $x^2=4ay$); and the circle $x=h+r\cos \theta$, $y=k+r\sin \theta$, centre $(h,k)$ and radius $r$.

Quick example. If $x=t-2$ and $y=t^2+1$, then $t=x+2$, so $y=(x+2{)}^2+1$: a parabola with vertex $(-2,1)$. Check any restriction on $t$, because it restricts $x$ and $y$ too (for example, $t\ge 0$ gives only the part with $x\ge -2$).

Cubic inequalities

When the cubic is already factored into linear factors, mark the zeros on a number line and sketch the cubic. A positive leading coefficient means the graph rises to the right. For example, $(x-1)(x+2)(x-3)\le 0$ is satisfied where the curve is on or below the axis: $x\le -2$ or $1\le x\le 3$.

Rational inequalities

Never multiply both sides by an expression like $(x+1)$ whose sign you do not know. Instead multiply by its square, which is positive, and remember the denominator cannot be zero.

Worked example 2: rational inequality

Solve $\frac{2x+5}{x+1}<3$.

Note $x\ne -1$. Multiply both sides by $(x+1{)}^2>0$:

Bring everything to one side and take out the common factor:

This parabola opens downwards with zeros $-1$ and $2$, so it is negative outside the zeros:

Check with a test value: $x=3$ gives $\frac{11}{4}<3$, true. $x=0$ gives $5<3$, false. Good.

Absolute value inequalities

$|x-a|$ is the distance from $x$ to $a$. So $|x-a|<k$ means "within $k$ of $a$" (one interval) and $|x-a|>k$ means "more than $k$ from $a$" (two rays).

Worked example 3: absolute value inequality

Solve $|3x+1|\ge 4$.

"At least 4 from zero" splits into two cases:

So $x\le -\frac{5}{3}$ or $x\ge 1$. Graphically, these are the $x$ values where the V-shaped graph $y=|3x+1|$ is on or above the line $y=4$.

• Multiplying a rational inequality by the denominator without squaring it, which flips the inequality for some $x$ values but not others.
• Including the value that makes the denominator zero in the answer (use a strict inequality there, even when the question has $\le$ or $\ge$).
• Writing $f^{-1}(x)=\frac{1}{f(x)}$. The inverse function is not the reciprocal.
• Taking the wrong sign of a square root when finding an inverse. Use the range of $f^{-1}$ (the domain of $f$) to decide.
• Confusing $y=|f(x)|$ (reflect the negative parts up) with $y=f(|x|)$ (copy the right half to the left).
• Forgetting that a parameter restriction (such as $0\le t\le 2$) restricts the Cartesian graph to part of a curve.
• Drawing the reciprocal graph through a point where $f(x)=0$ instead of putting an asymptote there.

• Sketch means a neat graph showing the key features: intercepts, asymptotes (as dashed lines with their equations) and turning points. Label them; markers look for them.
• For an inequality, show the critical values and how you decided which regions satisfy it (a sketch or a sign table). A correct answer with no reasoning may lose marks.
• When finding an inverse, state its domain. Many questions award a mark for the domain alone.
• Show that questions on inverses: start from the definition, for example verify $f(f^{-1}(x))=x$.
• In multiple choice graph questions, test one or two easy points (such as $x=0$, or where $f(x)=1$) to rule out options quickly.
• Check answers to inequalities by substituting a value from each region.

• Advanced, Working with Functions and Graph Transformations: the function language, quadratic inequalities, transformations and the graph of $y=|x|$ are assumed knowledge.
• Polynomials (ME11-PO): factored polynomials and their graphs feed directly into cubic inequalities.
• Further trigonometry (ME11-TR): reciprocal trig graphs and trig identities often appear together.
• Inverse trigonometric functions (ME12-IT): applies the inverse function ideas here, with domain restrictions, to $\sin$, $\cos$ and $\tan$.
• Introduction to vectors (ME12-VE): parametric equations reappear as the path of a projectile and of a particle moving in a straight line.