Monotonicity, Maxima and Minima of Functions | NIMCET

Quick answer: f is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$. Critical points are where $f'(x) = 0$ or f′ does not exist. By the second derivative test, at $f'(c) = 0$: if $f''(c) < 0$ there is a local maximum, and if $f''(c) > 0$ a local minimum. On a closed interval [a, b], the absolute maximum and minimum are among the critical values and the endpoint values $f(a)$, $f(b)$.

This follows tangents, normals and rates of change.

Monotonicity

A function can still be increasing if $f'(x) = 0$ at isolated points, as $x^3$ is at 0.

Finding extrema

Step 1: Find f′(x) and solve $f'(x) = 0$; also note points where f′ is undefined.

Step 2: Classify each critical point.

If $f''(c) = 0$, the second derivative test is inconclusive; use the first derivative test.

If f′ does not change sign at c, there is no extremum (for example $x^3$ at 0, a point of inflection).

Absolute extrema on [a, b]

Evaluate f at all critical points inside (a, b) and at a and b. The largest value is the absolute maximum and the smallest is the absolute minimum.

Standard optimisation results

Worked examples

Example 1: intervals of monotonicity

Find where $f(x) = x^3 - 3x^2 - 9x + 5$ is increasing.

Step 1: $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$.

Step 2: $f' > 0$ for $x < -1$ or $x > 3$. So f increases on $(-\infty, -1)$ and $(3, \infty)$, and decreases on $(-1, 3)$.

Example 2: local extrema

For the same function, $f''(x) = 6x - 6$.

• At $x = -1$: $f'' = -12 < 0$, local maximum $f(-1) = 10$.
• At $x = 3$: $f'' = 12 > 0$, local minimum $f(3) = -22$.

Example 3: absolute extrema

Find the absolute maximum and minimum of $f(x) = x^3 - 3x$ on $[0, 2]$.

Step 1: $f'(x) = 3x^2 - 3 = 0$ gives $x = 1$ in the interval.

Step 2: $f(0) = 0$, $f(1) = -2$, $f(2) = 2$.

Answer: maximum 2 at $x = 2$, minimum −2 at $x = 1$.

Example 4: optimisation

Find two positive numbers with sum 20 whose product is a maximum.

Step 1: $P = x(20 - x)$, $P' = 20 - 2x = 0$, so $x = 10$.

Step 2: $P'' = -2 < 0$, a maximum. The numbers are 10 and 10, with product 100.

Example 5: open box

From a 12 cm square sheet, equal squares are cut from the corners to make an open box. Find the side of the cut for maximum volume.

Step 1: $V = x(12 - 2x)^2$.

Step 2: $V' = (12 - 2x)(12 - 6x) = 0$ gives $x = 2$ (since $x = 6$ gives zero volume).

Answer: $x = 2$ cm, volume $2 \times 8^2 = 128$ cm³. (This matches $\frac{a}{6} = 2$.)

Summary table

Exam tips

• Always check endpoints for absolute extrema on a closed interval.
• In word problems, write the quantity in one variable before differentiating.
• Remember that a local maximum value can be smaller than a local minimum value elsewhere.

The AM-GM inequality solves many optimisation problems without calculus. See the NIMCET Calculus chapter for more.

Common mistakes

• Forgetting points where f′ is undefined.
• Ignoring endpoints when finding absolute extrema.
• Concluding there is an extremum wherever $f' = 0$.
• Confusing the x-value of an extremum with the extreme value.

Frequently Asked Questions

When is a function increasing?

On an interval where its derivative is positive.

What is the second derivative test?

At a critical point c with f′(c) = 0, f″(c) < 0 gives a local maximum and f″(c) > 0 gives a local minimum.

How do you find the absolute maximum on a closed interval?

Evaluate the function at the critical points and at the endpoints, and take the largest value.

What is a point of inflection?

A point where the curve changes concavity, such as x = 0 for y = x³.

What is the maximum product of two positive numbers with a fixed sum?

It occurs when the numbers are equal; for sum s, the maximum product is s²/4.