Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: complete function analysis (201-SN2-RE)

Q: How do you know whether a function has an oblique asymptote?

For a rational function, the degree of the numerator must exceed that of the denominator by exactly one. You then carry out the long division, and the polynomial part is the equation of the line. If the degree gap is two or more, there is no oblique asymptote but a branch that grows like a parabola.

Q: Does a denominator equal to zero always give a vertical asymptote?

No. The limit decides. If the factor cancels with the numerator, the limit is finite and the curve simply has a hole, not an infinite branch. You must therefore factor and simplify before studying the limits, or you will invent an asymptote that does not exist.

Q: Why does a zero derivative not always give an extremum?

Because an extremum requires the function to switch between increasing and decreasing, so the derivative must change sign. The cube of x has a zero derivative at zero, but that derivative stays positive on both sides: the curve keeps rising and forms an inflection point with a horizontal tangent, not an extremum.

Q: When does the second derivative test not work?

When the second derivative is zero at the critical number. The test becomes inconclusive exactly in the borderline cases, since the fourth power has a minimum and the cube has nothing, with the same zero second derivative. You must then go back to the sign chart of the first derivative, which always settles it.

Q: Can there be an inflection point on a vertical asymptote?

No. An inflection point is a point of the curve, so its x-coordinate must belong to the domain. The concavity often changes on either side of a vertical asymptote, but no point of the curve lies there, so there is no inflection point. This is a very common loss of marks at the end of an analysis.

This sheet does not repeat the list of steps of a function analysis: it is in your notes and nobody loses marks reciting it. It deals with what really costs marks, namely conclusions drawn too quickly from a correct calculation, and the final sketch that is never compared with the table you have just written.

It is written for CEGEP students in the Science program (Sciences de la nature) taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). The corrected exercise set on the same chapter then puts every reflex to the test.

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The thread of the chapter

A function analysis is not a list of calculations but an INVESTIGATION: every piece of information you find, the domain, an asymptote, the sign of $f'$ , the sign of $f''$ , must show up on the final sketch. A sketch that contradicts the table flags the error before the marker does.

This chapter is part of 201-SN2-RE, Differential Calculus

The essentials

The three asymptotes, and the condition for each

• VERTICAL at $x=a$ : a one-sided limit is infinite at $a$ . A zero denominator is a candidate, not a proof.
• HORIZONTAL: $\lim_{x\to\pm\infty}f(x)=L$ , finite. There can be TWO of them, one on each side.
• OBLIQUE: only if the degree of the numerator exceeds that of the denominator by EXACTLY one. It is found by division.
• Horizontal and oblique exclude each other on the same side: a branch cannot have both.
• A curve can CROSS its horizontal or oblique asymptote; it never crosses a vertical one.

\frac{x^{2}+1}{x-1}

has a vertical asymptote at

x=1

and an oblique asymptote

y=x+1

, obtained by division; there is NO horizontal asymptote, since the degrees rule it out.

Comparing the degrees before computing a single limit saves five minutes and stops you from looking for an oblique asymptote that does not exist.

The eight steps, in the order that avoids going back and forth

• 1. Domain. 2. Intercepts. 3. Symmetry: $f(-x)=f(x)$ or $-f(x)$ , periodicity.
• 4. Limits at the ends of the domain and at infinity, hence the asymptotes.
• 5. $f'$ , critical numbers, SIGN chart of $f'$ , hence increasing and decreasing behaviour and extrema.
• 6. $f''$ , zeros, sign chart of $f''$ , hence concavity and inflection points.
• 7. Variation table gathering everything.
• 8. Sketch, THEN check the sketch against the table.

Step 3 always pays off: an even or odd function cuts the work in half, and spotting it takes ten seconds.

What each derivative decides, and what it does not

• $f'>0$ : increasing. $f'<0$ : decreasing. $f'=0$ : a candidate, nothing more.
• $f''>0$ : concave up, the curve turns upward. $f''<0$ : concave down.
• Extremum at $c$ : $f'$ must CHANGE sign at $c$ .
• Inflection point at $c$ : $f''$ must CHANGE sign at $c$ , and $c$ must be in the domain.
• Second derivative test: $f'(c)=0$ and $f''(c)>0$ gives a local minimum, $f''(c)<0$ a maximum. If $f''(c)=0$ , the test does NOT conclude.

x^{3}-3x

f'

changes sign at

-1

(maximum) and at

1

(minimum);

f''=6x

changes sign at

0

, where the curve goes from concave down (orange) to concave up (blue).

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Looking for an oblique asymptote when the degrees rule it out

2 marks, and an invented line on the sketch

What not to write

“ $f(x)=\frac{x^{3}+1}{x-1}$ , I do the long division to find the oblique asymptote.”

What to write

“The degree of the numerator exceeds that of the denominator by TWO: there is no oblique asymptote, the curve grows like a parabola (a parabolic branch).”

\frac{x^{3}+1}{x-1}

behaves like

x^{2}

at infinity: the curve rises like a parabola, and no line goes along with it. A degree gap of two, so no oblique asymptote.

Why: The condition is a degree gap of exactly one. With two, the division gives a polynomial of degree two, which is not a line. Comparing the degrees BEFORE dividing is the reflex that settles the question in three seconds.

2. Concluding there is a vertical asymptote as soon as the denominator vanishes

2 marks, and a wrong sketch near the point

What not to write

“ $f(x)=\frac{x^{2}-4}{x-2}$ has a vertical asymptote at $x=2$ .”

What to write

“The factor cancels: $f(x)=x+2$ for $x\neq 2$ , so there is a HOLE at $(2, 4)$ , not an asymptote.”

\frac{x^{2}-4}{x-2}

equals

x+2

everywhere except at

2

: the curve is a line with a hole at

(2, 4)

, with no infinite branch. No vertical asymptote.

Why: A zero denominator is a candidate, never a proof. What decides is the LIMIT: infinite, it is an asymptote; finite, it is a hole. Not simplifying before studying the limit makes you invent an infinite branch where the curve is perfectly flat.

3. Taking $f'(c)=0$ for an extremum

2 marks in the table, and a sketch that wrongly turns back

What not to write

“ $f'(0)=0$ , so $f$ has an extremum at $0$ .”

What to write

“ $f'$ does not change sign at $0$ : there is no extremum.”

x^{3}

has a horizontal tangent at

0

, but keeps rising on both sides:

f'=3x^{2}

does not change sign, so there is no extremum.

Why: A critical number is a candidate. For $f(x)=x^{3}$ , $f'(x)=3x^{2}$ is positive on both sides of $0$ : the curve keeps rising. Only the SIGN chart of $f'$ settles it, and you have to write it.

4. Believing that $f''(c)=0$ gives an inflection point

2 marks, and a wrong concavity across the whole table

What not to write

“ $f(x)=x^{4}$ satisfies $f''(0)=0$ , so $0$ is an inflection point.”

What to write

“ $f''(x)=12x^{2}\ge 0$ does not change sign: the curve stays concave up and there is no inflection point.”

Two second derivatives equal to zero, two opposite conclusions:

x^{4}

stays concave up on both sides of

0

and has no inflection point, while

(x-3)^{3}

switches from concave down to concave up at

3

Why: This is the exact mirror of the previous pitfall, one level up. A zero of $f''$ is a candidate; an inflection point requires a CHANGE of sign. The counterexample $x^{4}$ is the one exams ask about, and it is worth remembering as is.

5. Placing an inflection point outside the domain

1 to 2 marks

What not to write

“ $f''$ changes sign at $x=1$ , so the curve has an inflection point at $1$ .”

What to write

“ $1$ does not belong to the domain of $f$ : the concavity changes on either side of the asymptote, but there is NO inflection point.”

The concavity changes on either side of

x=1

, but

1

is not in the domain: no point of the curve is there to be an inflection point.

Why: An inflection point is a POINT of the curve: it needs an $x$ -coordinate in the domain and a $y$ -coordinate. A change of concavity across a vertical asymptote is common and never produces an inflection point.

6. Applying the second derivative test when it does not conclude

the whole question on that critical number

What not to write

“ $f'(c)=0$ and $f''(c)=0$ , so there is no extremum.”

What to write

“ $f''(c)=0$ : the test does not conclude. I go back to the FIRST derivative test, which always settles it.”

x^{4}

has a genuine minimum at

0

, even though

f'(0)=f''(0)=0

: the second derivative test is silent, the first derivative test (

-

then

+

) settles it.

Why: The second derivative test is a shortcut, not a complete method: it is inconclusive precisely in the borderline cases. $x^{4}$ has a minimum at $0$ with $f''(0)=0$ , and $x^{3}$ has nothing at $0$ with the same $f''(0)=0$ . The sign chart of $f'$ tells them apart.

7. Forgetting to check that the sketch agrees with the table

from 2 marks to the whole question, depending on which of the two is wrong

What not to write

“The table says decreasing on $\left(1, 2.41\right)$ , but my sketch goes up there. No big deal.”

What to write

“My sketch contradicts the table: one of the two is wrong, and I look for the error before handing in.”

The table says “decreasing on

(1, 2.41)

”: the correct curve (green) goes down to the minimum. A sketch that rises from

x=1

(red) contradicts the table, and one of the two is wrong.

Why: The sketch is a free CHECK of the table, and the only one you have without a calculator. A contradiction spotted in time can be fixed; the same contradiction handed in to the marker costs twice, once on the table and once on the graph.

8. Ignoring symmetry and doing the work twice

no marks lost, but ten minutes of exam time

What not to write

“I study $f(x)=\frac{x^{2}}{x^{2}+1}$ on all of $\mathbb{R}$ , point by point.”

What to write

“ $f(-x)=f(x)$ : the function is EVEN, so I study it on $\left[0, +\infty\right)$ and complete it by reflection in the y-axis.”

\frac{x^{2}}{x^{2}+1}

is even: it is studied on

[0, +\infty)

(solid line) and the left half is obtained by reflection in the

y

-axis (dashed).

Why: An even or odd function cuts the analysis in half, and the long question of the exam is timed. Testing $f(-x)$ takes ten seconds and is done at step 3, before any derivative is computed.

Which method to choose

Which asymptote to look for, according to the degrees

For a rational function, compare the degree $n$ of the numerator with the degree $d$ of the denominator BEFORE any calculation.

If $n<d$ → horizontal asymptote $y=0$

Example: $\frac{x+1}{x^{2}+3}\to 0$
If $n=d$ → horizontal asymptote $y=\frac{a_{n}}{b_{d}}$ , the ratio of the leading coefficients

Example: $\frac{3x^{2}-1}{5x^{2}+x}\to\frac{3}{5}$
If $n=d+1$ → OBLIQUE asymptote, obtained by dividing the numerator by the denominator

Example: $\frac{x^{2}+1}{x-1}=x+1+\frac{2}{x-1}$ , so $y=x+1$
If $n>d+1$ → no asymptote of this kind: a parabolic branch or a branch of higher degree

Example: $\frac{x^{3}+1}{x-1}$ behaves like $x^{2}+x+1$

For a function with a radical or an exponential, comparing degrees does not apply: compute the two limits at infinity separately, because they can differ.

How to conclude about a critical number

You have solved $f'(c)=0$ . What follows decides what $c$ really is.

If $f'$ changes from $+$ to $-$ at $c$ → local MAXIMUM at $c$

Example: $f'(x)=-2x$ at $c=0$
If $f'$ changes from $-$ to $+$ at $c$ → local MINIMUM at $c$

Example: $f'(x)=2x$ at $c=0$
If $f'$ keeps the same sign on both sides → no extremum: it is a flat spot, often an inflection point with a horizontal tangent

Example: $f'(x)=3x^{2}$ at $c=0$ for $f(x)=x^{3}$
If you prefer the shortcut and $f''(c)\neq 0$ → $f''(c)>0$ gives a minimum, $f''(c)<0$ a maximum

Example: $f''(0)=2>0$ for $f(x)=x^{2}$
If $f''(c)=0$ → the second derivative test is INCONCLUSIVE: go back to the sign chart of $f'$

ABSOLUTE extrema on a closed interval are found among the critical numbers AND the two endpoints. On an open interval, the endpoints are not reached and another argument is needed.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A complete analysis, written up for the marker

When to use it: The long, heavily weighted question of the last 201-SN2-RE exam.

1 State the domain and the intercepts, in one line each.
2 Test for parity, and SAY so even when the answer is no: “ $f(-x)\neq\pm f(x)$ , no symmetry.”
3 Give the limits at the ends of the domain and at infinity, then state each asymptote with its equation.
4 Write $f'$ SIMPLIFIED and factored, give the critical numbers, then the sign chart.
5 Write $f''$ factored, give its zeros, then the sign chart and the inflection points with BOTH coordinates.
6 Sketch by placing the asymptotes first as dashed lines, then the key points, then the curve.

Concluding sentence

The curve has the vertical asymptote $x=1$ and the oblique asymptote $y=x+1$ ; it has a local maximum at $\left(1-\sqrt{2}, -0.83\right)$ and a local minimum at $\left(1+\sqrt{2}, 4.83\right)$ .

The trap: Giving an extremum by its x-coordinate alone. An extremum is a POINT: it needs both coordinates, and the marking scheme counts the y-coordinate separately.

Marking: Typically 1 mark per asymptote, 2 marks for the sign chart of $f'$ , 2 marks for that of $f''$ , 3 marks for a consistent sketch.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Compare the sketch with the table, line by line

Go back over each interval of the table and check on the drawing that the curve rises or falls as stated, and that it is concave up or concave down as stated.

The table states a decrease between $1$ and $1+\sqrt{2}$ ; if the sketched curve rises there, one of the two is wrong, and it is almost always a sign lost in $f'$ .
Check an oblique asymptote by the limit of the difference

Compute $\lim_{x\to\pm\infty}\left[f(x)-(mx+p)\right]$ : it must equal $0$ . This is the definition, and it is the only check that proves the line you found.

$\frac{x^{2}+1}{x-1}-(x+1)=\frac{2}{x-1}\to 0$ ✓. A difference that tends to a nonzero constant signals a division error.
Check $f'$ at a point where the variation is visible

Evaluate $f'$ at a point where the sketch clearly rises and check that the sign is positive.

At $x=5$ , the curve visibly rises; $f'(5)=\frac{25-10-1}{16}=\frac{14}{16}>0$ ✓.
Count the expected sign changes

A rational function has as many critical numbers as the numerator of $f'$ has real roots. A table that misses one has a missing column.

$f'(x)=\frac{x^{2}-2x-1}{(x-1)^{2}}$ : the quadratic has two roots, so two critical numbers, so a table with five columns counting the asymptote.

The typical problem, taken apart

Complete analysis of a function with an oblique asymptote

Analyse $f(x)=\dfrac{x^{2}+1}{x-1}$ : domain, asymptotes, increasing and decreasing behaviour, concavity, then sketch.

Step 1

Domain: $x-1\neq 0$ , so $\operatorname{dom}\,f=\mathbb{R}\setminus\{1\}$ . The numerator $x^{2}+1$ never vanishes, so the curve does not cross the $x$ -axis; it crosses the $y$ -axis at $f(0)=-1$ .

Why

Two lines that earn two marks and set the frame. Noticing that the numerator never vanishes saves you from looking for roots that do not exist.

Step 2

Parity: $f(-x)=\frac{x^{2}+1}{-x-1}\neq\pm f(x)$ , no symmetry. Degrees: $2$ in the numerator, $1$ in the denominator, a gap of ONE, so an oblique asymptote. Division: $\frac{x^{2}+1}{x-1}=x+1+\frac{2}{x-1}$ .

Why

Comparing the degrees replaces two limit calculations. The division gives both the asymptote $y=x+1$ and the remainder $\frac{2}{x-1}$ , which tells you on which SIDE the curve passes: above for $x>1$ , below for $x<1$ .

Step 3

Limits: $\lim_{x\to 1^{-}}f(x)=-\infty$ and $\lim_{x\to 1^{+}}f(x)=+\infty$ , because the numerator tends to $2>0$ and the denominator to $0$ while changing sign. Vertical asymptote $x=1$ .

Why

The two one-sided limits are asked for separately because they differ. A single limit written as $\pm\infty$ is worth half the marks: the sign is read from that of the denominator.

Step 4

$f'(x)=\frac{2x(x-1)-(x^{2}+1)}{(x-1)^{2}}=\frac{x^{2}-2x-1}{(x-1)^{2}}$ . The denominator is always positive, so the SIGN of $f'$ is that of the quadratic, whose roots are $1\pm\sqrt{2}$ , that is $-0.414$ and $2.414$ .

Why

Reducing the sign study to that of the numerator is the move that saves the most time on a rational function: a square in the denominator never changes the sign.

Step 5

Table: $f'>0$ on $\left(-\infty, 1-\sqrt{2}\right)$ , $f'<0$ on $\left(1-\sqrt{2}, 1\right)$ and on $\left(1, 1+\sqrt{2}\right)$ , $f'>0$ after that. Local maximum at $\left(1-\sqrt{2}, -0.83\right)$ , local minimum at $\left(1+\sqrt{2}, 4.83\right)$ .

Why

The local maximum is LOWER than the local minimum, which is impossible on a single interval but perfectly normal on either side of a vertical asymptote. Pointing it out proves you have read your own table.

Step 6

$f''(x)=\frac{4}{(x-1)^{3}}$ : negative for $x<1$ , positive for $x>1$ . The curve is therefore concave down to the left of the asymptote and concave up to the right. NO inflection point, since $1$ does not belong to the domain.

Why

This is the pitfall of step 6: the concavity really does change, but not at a point of the curve. Writing explicitly “no inflection point since $1\notin\operatorname{dom}\,f$ ” earns the mark that most papers lose.

The conclusion, written out

$f$ is defined on $\mathbb{R}\setminus\{1\}$ , has the vertical asymptote $x=1$ and the oblique asymptote $y=x+1$ , has a local maximum at $\left(1-\sqrt{2}, -0.83\right)$ and a local minimum at $\left(1+\sqrt{2}, 4.83\right)$ , is concave down on $\left(-\infty, 1\right)$ and concave up on $\left(1, +\infty\right)$ , with no inflection point at all.

The classic mistake on this problem: Announcing an inflection point at $x=1$ because $f''$ changes sign there. An inflection point belongs to the curve: it needs an $x$ -coordinate IN the domain, and $1$ is not in it.

Learn by heart

• Vertical asymptote: the LIMIT decides, not the zero denominator. A finite limit means a hole.
• Degrees: $n<d$ gives $y=0$ ; $n=d$ gives the ratio of the leading coefficients; $n=d+1$ gives an oblique asymptote; beyond that, none.
• Two horizontal asymptotes are possible, one on each side, especially with a radical.
• Extremum: $f'$ must CHANGE sign. Inflection point: $f''$ must CHANGE sign, AND the point must be in the domain.
• The second derivative test is inconclusive when $f''(c)=0$ : go back to the sign chart of $f'$ .
• An extremum is a POINT: x-coordinate AND y-coordinate.
• For a rational function, the sign of $f'$ is that of its NUMERATOR, since the square in the denominator is positive.
• Mandatory last step: check the sketch against the table.

Frequently asked questions

How do you know whether a function has an oblique asymptote?

For a rational function, the degree of the numerator must exceed that of the denominator by exactly one. You then carry out the long division, and the polynomial part is the equation of the line. If the degree gap is two or more, there is no oblique asymptote but a branch that grows like a parabola.

Does a denominator equal to zero always give a vertical asymptote?

No. The limit decides. If the factor cancels with the numerator, the limit is finite and the curve simply has a hole, not an infinite branch. You must therefore factor and simplify before studying the limits, or you will invent an asymptote that does not exist.

Why does a zero derivative not always give an extremum?

Because an extremum requires the function to switch between increasing and decreasing, so the derivative must change sign. The cube of x has a zero derivative at zero, but that derivative stays positive on both sides: the curve keeps rising and forms an inflection point with a horizontal tangent, not an extremum.

When does the second derivative test not work?

When the second derivative is zero at the critical number. The test becomes inconclusive exactly in the borderline cases, since the fourth power has a minimum and the cube has nothing, with the same zero second derivative. You must then go back to the sign chart of the first derivative, which always settles it.

Can there be an inflection point on a vertical asymptote?

No. An inflection point is a point of the curve, so its x-coordinate must belong to the domain. The concavity often changes on either side of a vertical asymptote, but no point of the curve lies there, so there is no inflection point. This is a very common loss of marks at the end of an analysis.

Practise it

Corrected exercises: Function analysis and curve sketching, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-curve-sketching. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: complete function analysis (201-SN2-RE)

The essentials

The three asymptotes, and the condition for each

The eight steps, in the order that avoids going back and forth

What each derivative decides, and what it does not

The mistakes that cost marks

1. Looking for an oblique asymptote when the degrees rule it out

2. Concluding there is a vertical asymptote as soon as the denominator vanishes

3. Taking f′(c)=0f'(c)=0f′(c)=0 for an extremum

4. Believing that f′′(c)=0f''(c)=0f′′(c)=0 gives an inflection point

5. Placing an inflection point outside the domain

6. Applying the second derivative test when it does not conclude

7. Forgetting to check that the sketch agrees with the table

8. Ignoring symmetry and doing the work twice

Which method to choose

Which asymptote to look for, according to the degrees

How to conclude about a critical number

How the answer is expected to be written

A complete analysis, written up for the marker

Check before you hand in

The typical problem, taken apart

Complete analysis of a function with an oblique asymptote

Learn by heart

Frequently asked questions

How do you know whether a function has an oblique asymptote?

Does a denominator equal to zero always give a vertical asymptote?

Why does a zero derivative not always give an extremum?

When does the second derivative test not work?

Can there be an inflection point on a vertical asymptote?

Corrected exercises: Function analysis and curve sketching, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

3. Taking $f'(c)=0$ for an extremum

4. Believing that $f''(c)=0$ gives an inflection point