Introduction to Derivatives · Grade 12

How do you find intervals of increase and decrease and points of extremum?

Differentiate, set the derivative equal to zero and solve to find the candidate points, then check the sign of the derivative on either side of each. Positive means increasing, negative means decreasing, and a change of sign at a point makes it a maximum or a minimum depending on the direction of the change.

Learning objectives

The order that always works

Differentiate. Set the derivative to zero and solve — these solutions are the only places where the behaviour can change. Then test the sign of the derivative in each interval they create.

The solving step is ordinary algebra, usually factorising or the quadratic formula, which is why grade 9 sits underneath this page.

Do not skip the sign test. A zero derivative marks a candidate, not a conclusion, and there is no way to classify it without looking on both sides.

Reading the table

Build a row of intervals separated by the candidate points and record the sign of the derivative in each. Positive means the function rises there; negative means it falls.

A change from positive to negative is a maximum: the function rose, levelled, then fell. Negative to positive is a minimum. No change of sign at all means the point is neither, even though the tangent is horizontal there.

Answering what was actually asked

An interval of increase is reported as a range of x values. A point of extremum is reported as a point, which means two coordinates — and the second comes from the function, not the derivative.

Read the question once more before writing the final line. Asked for the maximum value, give the y-coordinate; asked for where the maximum is, give the x. The two are different answers and both are sitting in your working.

Worked examples

  1. Find the intervals of increase and decrease of f(x) = x squared minus 6x

    1. Differentiate: 2x minus 6
    2. Set to zero: 2x minus 6 equals zero, so x equals 3
    3. Test to the left, say x = 0: the derivative is minus 6, negative
    4. Test to the right, say x = 4: the derivative is 2, positive

    Answer: Decreasing for x below 3, increasing for x above 3

  2. Find the extremum of that function and classify it

    1. The only candidate is x = 3
    2. The derivative goes from negative to positive there
    3. That is a minimum
    4. Complete the point: 9 minus 18

    Answer: A minimum at the point (3, −9)

  3. Find the candidate points of f(x) = x cubed minus 3x

    1. Differentiate: 3x squared minus 3
    2. Set to zero: 3x squared equals 3
    3. Divide by 3 and take the root, both signs

    Answer: x equals 1 and x equals minus 1

Common mistakes

Declaring a maximum from a zero derivative alone
A horizontal tangent can be a maximum, a minimum, or neither. Only the sign on either side decides, and for a function like x cubed the answer is neither.
Reporting an extremum with only one coordinate
A point of extremum is a point. The x comes from solving the derivative and the y from substituting into the function, and an answer missing one of them is incomplete.
Testing the sign by substituting into the function
It is the sign of the derivative that says whether the function rises or falls. Substituting into the function gives a height, which says nothing about direction.

What to remember

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