Essential function study in Year 5 of high school

At vocational schools (istituti professionali) the study of functions opens Year 5 between September and October, and the first test usually arrives between the end of October and the end of November — that is, five to ten weeks from now. Before going into the calculations, one thing has to be said clearly: at a vocational school maths is not a second-paper subject (the second written paper of the maturità, Italy’s school-leaving exam). It counts towards the average of grades that opens admission to the exam, and it serves to understand what a graph says — not for a science-high-school-style test. This post teaches the essential version, complete but minimal: no integrals, no theorems of differential calculus, and the second derivative at most as a side note, not as a topic for a test.

Why the essential version, and not the science high school one

If you study from a textbook designed for the science high school (liceo scientifico), the chapter on function study includes concavity, inflection points, the second derivative, existence theorems, cases with complicated roots and logarithms. On that syllabus you get frightened by topics that your test will never ask for. Here there are five steps, always the same, which are enough to build a correct graph and to answer what the teacher really asks: where the function is defined, where it is positive, what it does at the ends, where it increases.

If your problem, more than the complete graph, is limits and asymptotes taken calmly, this post on Year 5 at the languages high school covers the same initial block (domain, sign, limits) at a slower pace, stopping before the complete graph — here instead we go all the way to the first derivative and the finished drawing, with the angle of admission more than the second paper.

Step 1 — the domain

At vocational school the domain almost always reduces to three cases, to be checked one by one:

• denominator: if the function has a fraction, the denominator must be different from zero;
• even-index root: the argument under the root must be greater than or equal to zero;
• logarithm: the argument of the logarithm must be strictly greater than zero.

To solve these conditions you need basic inequalities: if the sign still gives you doubts, now is the time to go back to equations, inequalities and systems in Year 2, because from here on the domain and the sign of the function are the very same technique applied twice.

Step 2 — intercepts with the axes and sign

The intercept with the $y$ axis is found by calculating $f(0)$, when $0$ belongs to the domain. The intercepts with the $x$ axis are found by solving $f(x)=0$.

The sign of the function answers the question “for which $x$ is the graph above the axis and for which below?”. You solve the inequality $f(x)>0$ with the same method as the Year 2 inequalities: if the function is a fraction, you study the sign of the numerator and that of the denominator, and combine them with the sign rule. This step is not a detail to skip: it is what in the end lets you check whether the graph you have drawn is credible.

Step 3 — limits and asymptotes

At the points excluded from the domain (where the denominator vanishes) and at the ends of the domain (when it tends to $+\infty$ or $-\infty$) you calculate the limits.

• If the limit as $x$ tends to a value excluded from the domain is infinite, there is a vertical asymptote there.
• If the limit as $x\to \pm \infty$ is a finite number $L$, the line $y=L$ is a horizontal asymptote.
• If, for a fractional function, the degree of the numerator exceeds the degree of the denominator by exactly one, the function has no horizontal asymptote but has an oblique asymptote: you find it by dividing numerator by denominator and taking the part without remainder.

Nothing else is needed: no oblique asymptotes in more complicated cases, no fine distinction between types of indeterminate forms.

Step 4 — first derivative: increase and decrease

You calculate $f^{\prime }(x)$ and study its sign, with the same inequality method seen for the sign of the function in Step 2:

• where $f^{\prime }(x)>0$, the function increases;
• where $f^{\prime }(x)<0$, the function decreases;
• at the points where $f^{\prime }(x)=0$ (and the sign changes around it) there is a maximum or a minimum: if $f^{\prime }$ goes from positive to negative it is a maximum, if it goes from negative to positive it is a minimum.

You don’t need the second derivative to tell maxima from minima: it is enough to look at how the sign of $f^{\prime }$ changes as you cross the point.

Step 5 — the graph

At this point you put everything together: you draw the asymptotes as guide lines, you mark the intercepts with the axes, you respect the sign found in Step 2 (the curve is above the axis where the sign is positive, below where it is negative), and you make the curve rise or fall according to the sign of $f^{\prime }$ found in Step 4.

The consistency check is this: if in Step 2 you found that the function is positive in a certain interval, and in the drawing that part of the curve is below the $x$ axis, one of the two steps has a mistake. The right graph is the one in which all the steps tell the same story.

The most common mistake

The mistake I see most often is confusing the sign of the function (Step 2) with the sign of the derivative (Step 4), treating them as if they were the same information. They are not: the sign of $f$ says where the graph is above or below the $x$ axis; the sign of $f^{\prime }$ says where the graph rises or falls. A function can be negative and increasing at the same time — it is below the axis, but it is rising towards zero. Whoever mixes the two pieces of information draws graphs that look plausible but are wrong, and in a test loses marks even if every single calculation was right.

First worked exercise

Let’s study

Domain. The denominator vanishes at $x=1$, so the domain is $x\ne 1$.

Intercepts and sign. With the $y$ axis: $f(0)=\frac{-4}{-1}=4$. With the $x$ axis: $2x-4=0\Rightarrow x=2$.

For the sign I study numerator and denominator separately: $2x-4>0$ for $x>2$; $x-1>0$ for $x>1$. Combining with the sign rule,

Limits and asymptotes. For $x\to 1^{-}$ the denominator tends to $0$ from negative values and the numerator is about $-2$: the ratio tends to $+\infty$. For $x\to 1^{+}$ the denominator tends to $0$ from positive values: the ratio tends to $-\infty$. So $x=1$ is a vertical asymptote.

Numerator and denominator have the same degree (both first degree), so the asymptote is horizontal, with value equal to the ratio of the leading coefficients:

The horizontal asymptote is $y=2$.

First derivative. With the quotient rule:

The numerator is constant and positive, the denominator is a square so always positive (except that it is not defined at $x=1$): $f^{\prime }(x)>0$ for every $x$ in the domain. The function is always increasing, both to the left and to the right of the vertical asymptote, and has no maxima or minima.

The graph. To the left of $x=1$ the curve rises from a value just above $2$ (when $x\to -\infty$) up to $+\infty$ (near $x=1^{-}$); to the right of $x=1$ it rises from $-\infty$ (near $x=1^{+}$) up to a value just below $2$ (when $x\to +\infty$). The intercept $(2,0)$ and the fact that the function is negative between $1$ and $2$ are consistent with this behaviour: just after the asymptote the curve starts from very negative values, rises, crosses zero at $x=2$ and keeps climbing, becoming positive.

Before moving on to the second exercise, try redoing the sign of Step 2 on your own with the same inequality technique of Year 2: if you lack confidence there, practise with the Year 5 maths practice test on function study. It is calibrated on the liceo, so some questions go beyond your syllabus — skip the ones on integrals and on the second derivative, and concentrate on domain, sign, asymptotes and the first derivative, which are exactly the four steps just seen. The correct answer is free, the guided solution unlocks with your first lesson.

Second worked exercise

Let’s study

Domain. $x\ne 0$.

Intercepts and sign. The numerator $x^2+1$ never vanishes (it is always positive), so the function never touches the $x$ axis. The sign depends only on the denominator: $f(x)>0$ for $x>0$, $f(x)<0$ for $x<0$.

Limits and asymptotes. For $x\to 0^{+}$ the numerator tends to $1$ and the denominator to $0^{+}$: $f\to +\infty$. For $x\to 0^{-}$: $f\to -\infty$. Vertical asymptote $x=0$.

The numerator has degree $2$, the denominator degree $1$: the degree of the numerator exceeds that of the denominator by exactly one, so there is an oblique asymptote, not a horizontal one. I divide:

When $x\to \pm \infty$, the term $\frac{1}{x}$ tends to $0$, so the oblique asymptote is $y=x$.

First derivative. Writing $f(x)=x+\frac{1}{x}=x+x^{-1}$:

The denominator $x^2$ is always positive, so the sign of $f^{\prime }$ follows the numerator $x^2-1$: positive for $x<-1$ or $x>1$, negative for $-1<x<0$ or $0<x<1$.

So the function increases for $x<-1$ and for $x>1$, decreases for $-1<x<0$ and for $0<x<1$. At $x=-1$ the derivative goes from positive to negative: it is a maximum, with

At $x=1$ the derivative goes from negative to positive: it is a minimum, with

The graph. To the left of $x=0$: the curve comes from $-\infty$ following the asymptote $y=x$, rises to the maximum $(-1,-2)$, then falls towards $-\infty$ near the vertical asymptote. To the right of $x=0$: the curve comes down from $+\infty$ near the vertical asymptote, touches the minimum $(1,2)$, then climbs again following $y=x$ as $x\to +\infty$. The sign found earlier — negative to the left of zero, positive to the right — is consistent: the maximum $(-1,-2)$ is below the axis, the minimum $(1,2)$ is above.

To practise

Review the sign of inequalities with equations, inequalities and systems in Year 2 if Steps 2 and 4 were hard work, then go back to the Year 5 maths practice test on function study, skipping the off-syllabus questions. The rest of the maths year is in the Year 5 formula sheet.

If you are at a technical institute on the economics track, Year 5 on that track works on a completely different chapter — functions of two variables and constrained maxima — because even within the same “Year 5” the maths syllabus changes a lot from one track to another.

All the maths of the five years is collected at materie/matematica. If a part of this chapter is still unclear, the first chat is free → contact.