Critical points and the first derivative test

What you will be able to do

• say what a local maximum, a local minimum and a critical point are;
• find the critical points of a function;
• use the sign of $f'$ on both sides of a critical point to decide what it is;
• recognise a critical point that is not a maximum or a minimum.

Peaks and valleys

On a hill walk, the top of a small hill is higher than the ground near it, but there may be a higher hill elsewhere. A local maximum is such a top. In symbols, $f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ near $c$. It has a local minimum at $c$ if $f(c) \le f(x)$ for all $x$ near $c$. Together they are called local extreme values. Some books say relative maximum and minimum.

A turning point is a point of the graph where the direction changes from rising to falling or from falling to rising.

Critical points

A smooth hilltop has a flat tangent, so the gradient there is $0$. A sharp point, such as the tip of $|x|$, has no tangent at all. These two kinds of point are called critical points:

• points $c$ where $f'(c) = 0$, called stationary points in A-Level and IB;
• points $c$ in the domain where $f'(c)$ does not exist.

Only a critical point can be a local maximum or minimum. But not every critical point is one. The function $f(x) = x^3$ has $f'(0) = 0$, and it has no maximum or minimum at all.

The first derivative test

Walk along the graph from left to right and watch the sign of $f'$.

• If $f'$ changes from positive to negative at $c$, the graph rises and then falls. $c$ is a local maximum.
• If $f'$ changes from negative to positive at $c$, the graph falls and then rises. $c$ is a local minimum.
• If $f'$ keeps its sign, $c$ is neither. For $x^3$ we have $f'(0) = 0$, yet the graph only flattens for a moment and goes on climbing.

Take $f(x) = 2x^3 - 3x^2 - 12x + 5$. Then $f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1)$. The critical points are $x = -1$ and $x = 2$. The sign of $f'$ is $+$ before $-1$, $-$ between $-1$ and $2$, and $+$ after $2$. So $x = -1$ is a local maximum with $f(-1) = 12$, and $x = 2$ is a local minimum with $f(2) = -15$. Notice that the local maximum $12$ is not the greatest value of $f$: the graph climbs higher to the right.

A critical point where the derivative does not exist

$f(x) = |x|$ has no derivative at $x = 0$. To the left $f' = -1$ and to the right $f' = 1$. The sign changes from $-$ to $+$, so $x = 0$ is a local minimum. The same happens for $f(x) = x^{2/3}$, whose graph has a sharp tip at the origin.

On the other hand, $f(x) = x^{1/3}$ has no derivative at $0$ either, but $f' = \dfrac{1}{3x^{2/3}}$ is positive on both sides. So there is no extreme value at $0$.

A worked example

Find the local extreme values of $f(x) = x^4 - 2x^2$.

1. Differentiate. $f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$.
2. The critical points. $x = -1$, $x = 0$ and $x = 1$.
3. The signs. For $x < -1$: $f' < 0$. For $-1 < x < 0$: $f' > 0$. For $0 < x < 1$: $f' < 0$. For $x > 1$: $f' > 0$.
4. The result. Local minima at $x = \pm 1$ with $f(\pm 1) = -1$. Local maximum at $x = 0$ with $f(0) = 0$.

Practise with new numbers

Try it in Python

The program finds the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 5$. It looks at the sign of $f'$ just before and just after each one.

Common mistakes

• Treating every critical point as a maximum or a minimum. For $x^3$ at $0$ it is neither.
• Forgetting the points where $f'$ does not exist. Sharp tips such as $|x|$ at $0$ are critical points too.
• Testing the sign of $f$ instead of $f'$. The test is about the sign of the derivative on each side.
• Calling a local maximum the greatest value. A local value is the best only near the point.
• Skipping the test point on one side. Test one point on each side of the critical point.

Remember these

• Critical points: $f'(c) = 0$ or $f'(c)$ does not exist.
• Local extreme values can only be at critical points.
• $f'$ from $+$ to $-$: local maximum. From $-$ to $+$: local minimum. No change: neither.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the critical points of $f(x) = x^3 - 3x^2$.

2. The derivative of a function is $f'(x) = (x - 1)(x - 3)$. Classify its critical points.

3. Does $f(x) = x^3$ have a local maximum or minimum?

4. What kind of point does $f(x) = |x|$ have at $x = 0$?

5. Find the local maximum value of $f(x) = 6x - x^2$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on maxima and minima.
• OpenStax, Calculus Volume 1, Section 4.3 (Maxima and Minima) and Section 4.5 (Derivatives and the Shape of a Graph).