Do local extrema occur if and only if f'(x) = 0?

This is part of a series on common misconceptions.

True or False?

Local extrema of $f(x)$ occur if and only if $f'(x) = 0.$

Why some people say it's true: That is the first derivative test we were taught in high school.

Why some people say it's false: There are cases that are exceptions to this statement. Also, I was taught that this is just the first step of the first derivative test.

The statement is $\color{#D61F06}{\textbf{false}}$.

Knowing that a function's derivative is 0 at a specific point does not necessarily mean that there's a local minimum or maximum at that point. When $f'(x) = 0,$ it only implies that there is a horizontal tangent at that point. An example of the case when a function has a horizontal tangent at a point that is not a local extremum is given below.

Additionally, it's possible for a function to have a local minimum or maximum at a point where its derivative is non-zero in a variety of different ways outlined further below. For example, $f(x) = |x|$ has a local minimum at $x = 0,$ despite $f'(0) \neq 0.$

Counterexample 1:

This is a simple counterexample of the given statement: an example of when the derivative of a function is 0 at a point which is not a local extremum.

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Consider the example of $f(x) =x^3$, which is continuous and differentiable everywhere $($for all $x \in \mathbb{R}).$

Differentiating $f(x)$ with respect to $x,$ we have $f'(x) = 3x^2 .$ Setting it to zero, we get $x = 0.$ But graphically we can clearly see that the function does not attain a local extremum at $x = 0.$ The curve merely goes from being concave downwards for $x < 0$ to being concave upwards for $x > 0.$ $_\square$

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Counterexample 2:

$y = \sin x$ for $0 ≤ x ≤ \frac{\pi}{3}$ is an example of a function which has a local extremum at a point where its derivative is not 0. This occurs because the function has a restricted domain.

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Shown below is the graph of $y = \sin x$ for $0 ≤ x ≤ \frac{\pi}{3}.$

The function attains its maximum value at $x = \frac{\pi}{3}$ and minimum at $x = 0:$

$$\begin{array}{c}&f_{\text{max}}(x)=\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}, &f_{\text{min}}(x) = \sin 0 = 0.\end{array}$$

However, the derivative of the function $f'(x) = \cos x$ is not equal to zero at any of the points:

$$\begin{array}{c}&f '(0) =\cos 0 = 1, &f'\left(\dfrac{\pi}{3}\right) =\cos \dfrac{\pi}{3} = \dfrac{1}{2},\end{array}$$

implying the given statement is false. $_\square$

More information on local and global extrema can be found here: Extrema.

The following are two common objections to the counterexample given above:

The local extrema of $y = \sin x$ for $0 ≤ x ≤ \frac{\pi}{3}$ are

$$\begin{array}{c}&f_{\text{max}}(x)=\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}, &f_{\text{min}}(x) = \sin 0 = 0.\end{array}$$

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Rebuttal: The gradient of the function is positive at $x = \frac{\pi}{3}$, and therefore it is increasing and the value we got isn't the local maximum.
Reply: The function isn't defined for $x > \frac{\pi}{3}$ , and hence we have indeed reached the local maximum at $x = \frac{\pi}{3} .$

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Rebuttal: Isn't $-1$ the minimum value of $\sin x?$
Reply: The minimum value of the function $\sin x$ in the domain $0 ≤ x ≤ \frac{\pi}{3}$ is not $-1.$ The value $-1$ corresponds to the minimum value on a larger domain.

See Also

• List of Common Misconceptions
• Local Extrema
• Global Extrema
• Differentiation Rules