CEGEP Calculus I · Sheet 06 of 8 All 8 sheets →
  1. Home
  2. Worksheets
  3. CEGEP Calculus I
  4. Analysis of Functions and Curve Sketching
CEGEP Calculus I Applications of the Derivative Free · no sign-up

CEGEP Calculus I — Analysis of Functions and Curve Sketching

This is the set piece of the course. You are handed a rule and asked to produce everything its graph does — in order, with a reason on every line: where the function lives, where it meets the axes, what it does at the edges of its domain and out at infinity, where it rises and falls, where it bends, and finally the picture, which has to agree with all of it. No single step is hard. The marks go to whoever runs the same list every time and justifies each line from a derivative rather than from the shape they were expecting. Read what you like on this page, then print the free PDF when you want to write on it.

Practice worksheet — free PDF

5 pages 8 questions Letter size, print-ready

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 6 harder problems come with the CEGEP Calculus I bundle.

All 8 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

  1. Q1Domain, Zeros and Intercepts

    Determine the domain, the zeros and the y-intercept of each function. Where a feature does not exist, say so and explain why.

    1. f(x)=x34xx22x3
    2. h(x)=x+1lnx
  2. Q2Locating Vertical and Horizontal Asymptotes

    Let f(x)=3x212x2x6. Find every vertical and every horizontal asymptote of the graph of f, and justify each one with a limit. Identify the point where the denominator is zero but no vertical asymptote occurs, and give its coordinates.

  3. Q3Increasing and Decreasing Intervals

    Let f(x)=x2ex. Find f, then determine every interval on which f is increasing and every interval on which it is decreasing, supporting your answer with a sign analysis of f.

  4. Q4Relative and Absolute Extrema

    Let f(x)=2x39x2+12x on the closed interval [0,3].

    1. Find the critical numbers of f and classify each one with the first derivative test.
    2. Find the absolute maximum and the absolute minimum of f on [0,3], stating where each occurs.
  5. Q5Concavity and Points of Inflection

    Let f(x)=x44x3+5. Determine the intervals on which the graph of f is concave up and those on which it is concave down, and give the coordinates of every point of inflection.

  6. Q6Sketching a Complete Curve

    Carry out a complete study of f(x)=x2x24 — domain, intercepts, symmetry, vertical and horizontal asymptotes with the relevant limits, intervals of increase and decrease with any relative extrema, and intervals of concavity with any points of inflection — then sketch the curve on the grid, labelling each asymptote.

    A blank Cartesian grid for this question is on the printable PDF.

  7. Q7Rolle's Theorem and the Mean Value Theorem

    Name each hypothesis as you check it.

    1. Verify that f(x)=x34x satisfies the hypotheses of Rolle's Theorem on [2,2], then find every value of c the theorem guarantees.
    2. Verify that g(x)=x satisfies the hypotheses of the Mean Value Theorem on [1,9], then find every value of c the theorem guarantees.
  8. Q8Synthesis — drawing on several topics in this set

    Let f(x)=6xx2+3. Carry out a full analysis:

    1. domain, zeros, y-intercept and symmetry;
    2. every vertical and horizontal asymptote, justified by a limit;
    3. intervals of increase and decrease, and every relative extremum;
    4. the absolute maximum and absolute minimum values of f on , with a reason — note that the Extreme Value Theorem does not apply here;
    5. intervals of concavity and every point of inflection;
    6. a short description of the resulting shape.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This sheet is built for CEGEP Calculus I — 201-NYA-05 under the legacy numbering, 201-SN2-RE under the current one — and it follows the ministerial devis for competency 0M02, whose study-of-a-function criterion names exactly what is here: domain, zeros and the y-intercept; vertical and horizontal asymptotes; intervals of increase and decrease with relative and absolute extrema; intervals of concavity and points of inflection; and the sketch. Rolle's Theorem and the Mean Value Theorem are on the sheet as well, because the competency asks for rigorous reasoning from named theorems. Antiderivatives, the substitution rule and differential equations are deliberately absent: the devis places those in Calculus II, not here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The ministerial list is the method

The devis does not ask for a drawing. It asks for an étude d'une fonction, and it names the parts in order. Q8 is that whole list carried out on a single function; Q1 to Q7 are its parts practised one at a time. Run the list in the same order on every function you are given and the sketch stops being an act of drawing — it becomes a summary of what you have already proved.

The study of a function, in the order the devis lists it

Each stage feeds the next. Skip one and a later stage has nothing to stand on.

  1. 1
    Domain

    Every statement you make afterwards is a statement about points of the domain. Q1 is on the sheet to make this the reflex it needs to be.

  2. 2
    Zeros and the y-intercept

    A quotient is zero where its numerator is zero and the quotient is defined; a product is zero where a factor is zero and the product is defined. The y-intercept is f(0) — when 0 is in the domain at all.

  3. 3
    Symmetry, if it is free

    Test f(−x) once. Even or odd halves the work in Q6 and Q8, and it is a check on the finished sketch.

  4. 4
    Vertical and horizontal asymptotes, each justified by a limit

    Q2 asks for the limits explicitly, and that is the standard: an asymptote is a claim about a limit, so the limit is the evidence.

  5. 5
    The sign of f’: increase, decrease, extrema

    Q3 and Q4. One sign chart answers increasing, decreasing and every relative extremum at once.

  6. 6
    The sign of f’’: concavity and points of inflection

    Q5. The same machine, with a different function fed into it.

  7. 7
    The sketch

    Q6 and Q8. Asymptotes dashed, the computed points plotted and labelled, arcs bending the way the second chart says.

Domain first, because everything after it lives inside the domain

Q1 looks like the easy question on the sheet, and it is the one that quietly decides the rest. A rational rule excludes the zeros of its denominator. A square root needs its radicand to be at least zero. A logarithm needs its argument strictly positive. When a rule combines two of these — as the second function in Q1 does — the conditions have to hold at the same time, so you intersect them rather than list them.

Then filter everything through it. Solve for the zeros in the usual way, and throw out any solution that is not in the domain: a factor can vanish at a value the function never reaches, and that value is not a zero of the function. The y-intercept gets the same treatment — it exists only if 0 is in the domain, and a graph has at most one of them.

Q1 tells you to say so and explain why when a feature does not exist. That instruction is not padding. “No y-intercept, because 0 is not in the domain” is a complete answer; a blank space is not.

A zero of the denominator is not automatically a vertical asymptote

This is the whole point of Q2, which asks you to find the value where the denominator vanishes and no asymptote appears. Factor the numerator and the denominator before you conclude anything, and cancel what cancels — while noting that the cancelled value stays out of the domain regardless.

denominator zero + infinite one-sided limit → vertical asymptote; denominator zero + finite limit → a hole

Vertical. For each surviving zero of the denominator, take the limit from the left and from the right separately. The sign of the numerator there, together with the side the small denominator approaches zero from, gives you the direction. The two sides need not agree, and Q2 wants both written down.

Horizontal. Divide numerator and denominator by the highest power of x in the denominator, then take the limit. A horizontal asymptote is a statement about end behaviour only — it says nothing about the middle of the graph, and a curve is perfectly entitled to cross one.

Critical numbers: the half of the definition that gets dropped

A critical number of f is a value c in the domain of f at which either f’(c) = 0 or f’(c) does not exist. Setting the derivative equal to zero finds some critical numbers. It does not find all of them.

Where f’ fails to exist. Once your derivative is written as a single fraction, ask two questions of it rather than one: where is the numerator zero, and where is the denominator zero? A value of the second kind that is still in the domain of f is a critical number you are required to list — that is how cusps, corners and vertical tangents arrive, and they are very often exactly where the extremum sits.

The mirror trap. In Q6 the derivative also fails to exist at the two values excluded from the domain of f. Those are not critical numbers, because they are not in the domain — they are the vertical asymptotes. They still split the number line, though: a sign chart is broken at every point where the expression can change sign, whether or not the function is defined there. Two different rules, both worth saying out loud as you build the chart.

Relative is a local claim; absolute is a claim about everything

Q4 separates the two on purpose. Part (a) classifies the critical numbers with the first derivative test — f’ changing from positive to negative gives a relative maximum, negative to positive gives a relative minimum, and no sign change gives neither. Part (b) then asks a different question on a closed interval, and the answer to (a) does not settle it.

The endpoints are candidates, and nothing in your sign chart will remind you of them. On a closed interval [a, b] with f continuous, the Extreme Value Theorem promises that an absolute maximum and an absolute minimum exist. Each occurs either at a critical number inside the interval or at an endpoint — so evaluate f at every critical number in [a, b] and at both a and b, then compare the values. A relative maximum found in part (a) is beaten by an endpoint far more often than students expect, and the only way to know is to compute f(a) and f(b).

Then give both halves of the answer: the absolute maximum value, and the x at which it occurs. A bare x is not a maximum value, and a bare number is not a location.

Q8 asks for the absolute extrema on all of the reals, where the Extreme Value Theorem has nothing to say — its hypothesis of a closed, bounded interval fails, and the question tells you so. The argument there comes from the monotonic behaviour instead: if f decreases on one side, rises to a value, and decreases after it, then no value of f can exceed that one. That is a complete proof, and it needs no theorem to license it.

Concavity, inflection, and the asymptote that fakes one

Q5 is the second sign chart, run on f’’. Concave up where f’’ > 0, concave down where f’’ < 0, and a point of inflection where the sign changes. A zero of f’’ is only a candidate: it marks where to look, and the sign change is the fact.

A point of inflection is a point of the graph. This is where Q6 catches people. The sign of f’’ changes as you cross a vertical asymptote — of course it does, the curve is on two separate branches there — but the function is not defined at that value, so there is no point of the graph available to be an inflection point. The change of concavity is produced by the asymptote, not by a bend in the curve.

Two conditions, both required: c lies in the domain of f, and f’’ changes sign at c. Then report the answer as coordinates, because a point of inflection is a point — which means going back and evaluating f(c).

Rolle and the Mean Value Theorem: name each hypothesis as you check it

Q7 says so in its instruction, because at this level the check is the answer. A theorem applies when its hypotheses hold, and writing down which ones hold and why is the rigorous reasoning the competency asks for.

Rolle's Theorem. f continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b). Then some c in (a, b) has f’(c) = 0. For a polynomial the first two clauses are one sentence; the third has to be computed.

Mean Value Theorem. The same first two hypotheses, with the equal-endpoint condition dropped. Then some c in (a, b) has f’(c) equal to the average rate of change [f(b) − f(a)] / (b − a) — the slope of the secant through the two endpoints.

The interval you check on is not the domain. Q7 hands you a square root, whose derivative fails to exist at the left end of its own domain. If that value is not in the interval you were given, the failure is irrelevant — but say so, rather than passing over it in silence.

Every c the theorem guarantees. Solve the equation completely, then keep only the solutions lying in the open interval. There may well be more than one, and a solution landing exactly on an endpoint is rejected.

The sketch has to agree with the work above it

Q6 and Q8 both end in a curve, and the curve is judged against the analysis rather than against how smooth it looks. Draw every asymptote dashed first, then plot only the points you actually computed — intercepts, relative extrema, points of inflection — label them with their coordinates, and join them with arcs that bend the way your second sign chart says they do.

Read the sketch back as a test. A curve that turns around inside an interval where your chart says f’ keeps one sign is wrong. A curve that crosses a vertical asymptote is wrong. A curve running off in the same direction on both sides of an asymptote contradicts one-sided limits that disagreed. Each of those errors is visible in two seconds and invisible in the algebra.

Near a horizontal asymptote, decide whether the curve settles onto it from above or from below — substitute one large value of x if you are unsure. That detail is part of the end behaviour, and it is the difference between a sketch and a shape.

Getting the most out of it

Run the list, in the same order, every time

Domain, zeros and intercept, symmetry, asymptotes, f’, f’’, sketch. Write those headings down your page before you compute anything, and fill them in. The order is not a preference — it is the order in which each step supplies what the next one needs, and it is the order the devis lists.

Factor before you conclude anything

Almost every trap in this material is disarmed by factoring first: the difference between a hole and a vertical asymptote, the zeros that turn out not to be in the domain, the sign of a derivative you would otherwise have to multiply out. A factored expression hands you its zeros and the sign of each piece for nothing.

Justify from the derivative, not from the picture

“f is increasing on (0, 2) because f’(x) > 0 there” earns the mark; “the graph goes up” does not. When you use a named theorem, name it and check its hypotheses in the same breath. When a theorem does not apply, say which hypothesis fails and where.

Ask whether the interval is closed, every single time

Closed and bounded: the Extreme Value Theorem applies, the extrema are guaranteed to exist, and the endpoints join the list of candidates. Open, or all of the reals: no guarantee, no endpoints, and you argue from monotonic behaviour or from limits instead. A large share of the avoidable marks lost in this material is the wrong branch of that one question.

Finish by checking the sketch against your own sentences

Read your conclusions back one at a time and look at the curve while you do it. Decreasing here, concave up there, this asymptote approached from above. Anything the picture and the prose disagree about is an error you can still fix.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus I Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Analysis of Functions and Curve Sketching

Three PDFs · 13 pages · all three are in the bundle below.

  • Answer key — 5 pages. All 8 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
  • Challenge problems — 5 pages, 6 problems. A separate sheet at exam-plus difficulty covering the same 7 concepts. Harder than anything on the free sheet.
  • Challenge answer key — 3 pages. Every challenge problem worked to the same standard, with the checks shown.
  • PDF, letter size, print-ready.
In the bundle See what's in it Not sold separately

The one thing that's for sale

Best value for the whole year

Every CEGEP Calculus I topic — the complete Solutions Bundle

One download, one payment, the whole program. Every answer key and every challenge set for all 8 CEGEP Calculus I worksheet sets — including this one.

8 sets · 24 PDFs · 82 pages$19.99
  • Worked solutions, not answer lists — every step written out
  • Covers the whole year's program at this level
  • Less than the price of one hour of tutoring — for the entire year's solutions
Everything paid, in one file $19.99CAD · one payment CEGEP Calculus I bundle — coming soon Not on sale yet

Taking Secondary 1 Math as well? The Secondary 1 Math bundle covers all 15 of its sets — 45 PDFs, 182 pages — on the same terms.

Taking Secondary 2 Math as well? The Secondary 2 Math bundle covers all 14 of its sets — 42 PDFs, 181 pages — on the same terms.

Taking Secondary 3 Math as well? The Secondary 3 Math bundle covers all 11 of its sets — 33 PDFs, 154 pages — on the same terms.

Taking Secondary 4 Math as well? The Secondary 4 Math bundle covers all 17 of its sets — 51 PDFs, 154 pages — on the same terms.

Taking Secondary 5 Math as well? The Secondary 5 Math bundle covers all 21 of its sets — 63 PDFs, 228 pages — on the same terms.

Taking CEGEP Calculus II as well? The CEGEP Calculus II bundle covers all 8 of its sets — 24 PDFs, 90 pages — on the same terms.

Taking CEGEP Linear Algebra as well? The CEGEP Linear Algebra bundle covers all 7 of its sets — 21 PDFs, 82 pages — on the same terms.

Taking AP Calculus AB as well? The AP Calculus AB bundle covers all 8 of its sets — 24 PDFs, 126 pages — on the same terms.

Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete Calculus I solutions bundle, which covers every set at this level.

Which CEGEP course is this for?

Calculus I — 201-NYA-05 under the legacy numbering and 201-SN2-RE under the current one, the same course under two codes. It is the differential calculus course in Sciences de la nature, and this set covers the study of a function under competency 0M02.

Does it follow the ministerial devis?

Item by item. The 0M02 criterion on applying the methods of differential calculus to the study of a function lists domain, zeros and the y-intercept; vertical and horizontal asymptotes; intervals of increase and decrease with relative and absolute extrema; intervals of concavity and points of inflection; and the sketch. Those are the sections of this sheet, in that order. Rolle's Theorem and the Mean Value Theorem are added because the competency requires reasoning from named theorems.

What exactly counts as a critical number?

A value c in the domain of f where f'(c) = 0 or where f'(c) does not exist. Both halves count. A derivative written as a fraction fails to exist wherever its denominator vanishes, and if that value is still in the domain of f it is a critical number you have to list — cusps, corners and vertical tangents all arrive that way. A value excluded from the domain is not a critical number, however badly the derivative behaves near it.

Why is a zero of the denominator not always a vertical asymptote?

Because the test is the limit, not the denominator. If the factor causing the trouble cancels against the numerator, the limit at that value is finite and the graph has a hole there — a single missing point, with the value still excluded from the domain. A vertical asymptote requires the one-sided limit to be infinite, which is why the justification for one is always a limit.

Do I really have to check the endpoints of a closed interval?

Yes, and skipping them is the most common way to lose the second half of a question. An absolute maximum has to beat every value on the interval, and the endpoints are not critical numbers, so no amount of derivative work will surface them. Evaluate f at each critical number inside the interval and at both endpoints, then compare. The endpoint wins more often than you would guess.

f''(c) = 0. Is that a point of inflection?

Not on its own. A point of inflection needs c to be in the domain of f and needs f'' to change sign at c. A zero of f'' where the sign does not flip is not an inflection point, and a sign change across a vertical asymptote is not one either — there is no point of the graph there to be one. Find the candidates from the zeros of f'' and the values where it is undefined, then test the sign on each side.

Where are antiderivatives and differential equations?

In Calculus II. The devis assigns integration, the substitution rule and differential equations to the next competency, so they are out of scope here and open Calculus II instead.

Can teachers use this in class?

Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

I'm stuck on one question. Can you help?

Yes — through one-on-one tutoring, in Montreal or online. Get in touch to arrange a session, or see the current rates.

← All 8 CEGEP Calculus I worksheets  ·  Secondary 1 Math series (15 sheets) →  ·  Secondary 2 Math series (14 sheets) →  ·  Secondary 3 Math series (11 sheets) →  ·  Secondary 4 Math series (17 sheets) →  ·  Secondary 5 Math series (21 sheets) →  ·  CEGEP Calculus II series (8 sheets) →  ·  CEGEP Linear Algebra series (7 sheets) →  ·  AP Calculus AB series (8 sheets) →

Download the free worksheet

Ready to improve your grades?

WhatsApp is the way to reach me — tell me the course you're taking and what you're stuck on, and we'll sort out a first session from there.

Message Me on WhatsApp

or send a message

I reply within a day, usually sooner. Your details are used only to answer you — see the Privacy Policy.

Private math & science tutoring in Montreal, QC — Westmount · Outremont · Town of Mount Royal · Hampstead · Côte-Saint-Luc · NDG · Nuns' Island · West Island — and online across Quebec.

Chat with Marius