McGill University · MATH 140

McGill MATH 140: Calculus 1 tutoring

Review functions, limits, continuity, derivatives and antiderivatives for McGill MATH 140, Calculus 1. Try an explained exercise or get one-to-one help choosing a method and understanding each step. You can start with the question alone.

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Course overview

McGill MATH 140 introduces functions and graphs, limits, continuity, derivatives, antiderivatives and applications. You learn how to differentiate functions and find antiderivatives, building on the function skills used later in integral calculus.

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MATH 140 topics to review

Functions and graphsLimitsContinuityDerivativesDifferentiation methodsAntidifferentiation and applications

Common difficulties

  • Choosing a derivative rule for a composite function
  • Setting up optimization and related-rates models

MATH 140: a short limits and derivatives diagnostic

First make a function continuous and find its derivative from the definition. Then combine derivative rules to find a tangent line.

Add antiderivatives and applications if they are on your review list.

Before you start: Allow 15 minutes for both questions, then 15 minutes to review the step that gave you trouble.

1. Fill a hole without changing nearby behaviour

For x≠2x\ne2, let g(x)=(x2−4)/(x−2)g(x)=(x^2-4)/(x-2), and set g(2)=kg(2)=k. Find the value of kk that makes gg continuous at 22. For that value, find g′(2)g^{\prime}(2) from a difference quotient.

Show the explained solution
  1. For x≠2x\ne2, factor and cancel to get g(x)=x+2g(x)=x+2. Thus lim⁡x→2g(x)=4\lim_{x\to2}g(x)=4. Continuity requires the separately assigned value k=4k=4.
  2. With k=4k=4 and h≠0h\ne0, g(2+h)−g(2)h=(4+h)−4h=1.\frac{g(2+h)-g(2)}{h}=\frac{(4+h)-4}{h}=1. Taking the limit gives g′(2)=1g^{\prime}(2)=1.
  3. Cancelling a factor did not assign the missing value by itself. The chosen value first makes the whole function agree locally with a line; that allows the derivative calculation.

Repair cancelled factors, domains and limits

2. Combine product and chain rules for a tangent

Let f(x)=x2e−xf(x)=x^2e^{-x} for real xx. Find f′(x)f^{\prime}(x) and the tangent line at x=1x=1.

Show the explained solution
  1. The outer operation is a product. The exponential also needs the chain rule: (e−x)′=−e−x(e^{-x})^{\prime}=-e^{-x}. Therefore f′(x)=2xe−x−x2e−x.f^{\prime}(x)=2xe^{-x}-x^2e^{-x}.
  2. At x=1x=1, both f(1)f(1) and f′(1)f^{\prime}(1) equal 1/e1/e. Point-slope form gives y−1e=1e(x−1),y-\frac1e=\frac1e(x-1), or y=x/ey=x/e.
  3. Check that the line passes through the original point. The slope alone does not determine the tangent line; omitting the inner minus sign gives the wrong slope.

Repair derivative-rule selection

Use your answers to plan your revision
  1. Start with the first incorrect line

    For a domain or limit error, use the cancelled-factors lesson linked in question 1 and explain which equality needs an excluded input. For a derivative error, use the three-question derivative-rule check linked in question 2.

  2. Extend to an application

    After the diagnostic, use the open-box worked example on this page. Cover its solution, write the physical domain, then compare the admissible stationary point with the boundary behaviour.

  3. Finish with mixed practice

    Choose matching questions from the limits-and-derivatives practice linked below. It was organized around Concordia MATH 203 topics; use only the questions matching your McGill outline and add your assigned antiderivative work if it is in scope.

If you would like help with these questions, you can start a lesson with a limit, derivative or application question you could not begin. If you have an attempt, your tutor can review it with you; otherwise, you can work through the first step together and clarify the algebra, definition or derivative rule you need.

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Course topic sources

Preparing for this course

Review high-school calculus, function graphs, factoring, trigonometric identities and exponentials/logarithms. MATH 139 is listed as not offered in the 2026–27 departmental catalogue. Ask McGill advising about a currently available route if your calculus background is limited; MATH 139 and 140 are not a required sequence.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with McGill University.

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Antiderivatives: use one value to find the function

A differentiable function satisfies f′(x)=3x2−4x+2f^\prime(x)=3x^2-4x+2 for every real xx and f(0)=5f(0)=5. Find f(x)f(x).

Before calculating, decide whether the derivative alone determines a single function. What additional information does f(0)=5f(0)=5 provide?

Show a hint

Find one term whose derivative is 3x23x^2, one whose derivative is −4x-4x and one whose derivative is 22. A constant disappears when differentiated.

Show the worked solution
  1. Reverse the power rule term by term. Every antiderivative on the real line has the form f(x)=x3−2x2+2x+Cf(x)=x^3-2x^2+2x+C, where CC is a constant.

  2. The derivative does not choose CC. The initial value does: f(0)=C=5f(0)=C=5. Therefore f(x)=x3−2x2+2x+5f(x)=x^3-2x^2+2x+5.

  3. Check both requirements. Differentiation recovers 3x2−4x+23x^2-4x+2; substituting x=0x=0 gives 55. Checking only the derivative would not detect an incorrect constant.

Retry without the solution

Find g(x)g(x) if g′(x)=6x2−4g^\prime(x)=6x^2-4 for every real xx and g(1)=3g(1)=3. Check the derivative and the given value separately.

Check the retry

An antiderivative is g(x)=2x3−4x+Cg(x)=2x^3-4x+C. Now g(1)=2−4+C=3g(1)=2-4+C=3, so C=5C=5. Thus g(x)=2x3−4x+5g(x)=2x^3-4x+5. Its derivative is 6x2−46x^2-4 and g(1)=3g(1)=3. The given function value is not automatically the constant; substitute its input for xx before solving for CC.

If checking the derivative is difficult, use the derivative-rule check below. If the derivative is right but the initial value fails, return to the substitution used to find CC. A tutor can ask you to explain both checks, then change the initial value for a new attempt.

Optional refresher: expand a difference quotient

Let f(x)=x2f(x)=x^2 for real input xx, and let h≠0h\ne0 be a change from the input 3. Simplify f(3+h)−f(3)h\frac{f(3+h)-f(3)}{h}.

Show the reasoning

Expand the numerator: (3+h)2−9=6h+h2=h(6+h)(3+h)^2-9=6h+h^2=h(6+h). Since h≠0h\ne0, cancellation gives 6+h6+h. This is the average rate of change of the function over that change in input. Its limit as h→0h\to0 is 66, the instantaneous rate of change given by the derivative f′(3)f^\prime(3).

An optimization answer must fit the physical domain

Finding where a derivative is zero is only one step in an optimization problem. This box example uses geometry to find the allowed dimensions, then checks which choice gives the largest volume.

1. Build the model from the dimensions

Write V(x)V(x) for the box’s volume in cubic centimetres when the corner cut has side xx centimetres. Cut a square of side xx centimetres from each corner of a 30 cm×20 cm30\text{ cm}\times20\text{ cm} sheet and fold up an open box. Its height is xx, and its base measures 30−2x30-2x by 20−2x20-2x. Therefore V(x)=x(30−2x)(20−2x).V(x)=x(30-2x)(20-2x). The height and both base dimensions must be positive, so the allowed cut sizes form the physical domain 0<x<100<x<10.

Cut a sheet to make an open boxA 30 by 20 centimetre sheet with four squares of side x removed. The base is 30−2x by 20−2x.30 cm20 cmxxxx30 − 2x× (20 − 2x)Dashed lines: folds
Remove four equal squares of side xx, then fold the outer strips upward. The box height is xx; keeping its height and both base dimensions positive requires 0<x<100<x<10 cm.

2. Differentiate the quantity being optimized

Expanding gives V(x)=600x−100x2+4x3,V′(x)=600−200x+12x2.\begin{gathered}V(x)=600x-100x^2+4x^3,\\V'(x)=600-200x+12x^2.\end{gathered} Setting V′(x)=0V'(x)=0 yields x=25±573.x=\frac{25\pm5\sqrt7}{3}. The roots are approximately 3.9243.924 and 12.74312.743. The larger one is outside the physical domain and cannot describe this box.

3. Show why the remaining candidate is a maximum

Write x∗x_* for the remaining allowed cut size where the derivative is zero; the subscript is just a name for this particular cut size. The polynomial model extends continuously to [0,10][0,10], where both endpoint volumes are zero. There is only one point inside this interval where the derivative is zero, x∗=25−573.x_* =\frac{25-5\sqrt7}{3}. Its volume is positive, so comparing this candidate with the endpoints establishes the global maximum. Equivalently, V′V' changes from positive to negative there.

4. Interpret the answer with units

Cut squares of side about 3.924 cm3.924\text{ cm}. The maximum volume is approximately V(x∗)≈1056.31 cm3.V(x_*)\approx1056.31\text{ cm}^3. A length is the decision variable; a volume is the optimized outcome. Keep those units separate and round only after evaluating the exact expression.

Try it yourself

Repeat the model for a square sheet measuring 24 cm×24 cm24\text{ cm}\times24\text{ cm}. What corner cut maximizes the open box’s volume?

Show the explained solution

Here V(x)=x(24−2x)2V(x)=x(24-2x)^2 with 0<x<120<x<12, and V′(x)=(24−2x)(24−6x).\begin{gathered}V'(x)\\=(24-2x)(24-6x).\end{gathered} The interior candidate is x=4 cmx=4\text{ cm}; x=12x=12 is an endpoint where the base dimensions become zero. Comparing with the zero endpoint volumes gives the maximum 4⋅162=1024 cm34\cdot16^2=1024\text{ cm}^3.

On another optimization problem, write the physical domain before differentiating. Explain how each candidate is checked against boundaries and the original question.

How tutoring can help

You can start MATH 140 tutoring with a question you have not been able to begin. Your tutor can explain the relevant algebra, show how to read a limit and help you choose a derivative rule. For an application, you can identify the quantities and build the equation together. For optimization, your tutor can help you find the allowed values and justify a maximum or minimum. For an antiderivative, you can check the answer together by differentiating it, then try a similar question with guidance.

Exam preparation for MATH 140

Practise derivative selection and interpretation separately. For optimization, compare valid candidates and endpoints. Include basic antiderivatives where your outline requires them.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

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