How Do You Find Local Maximum and Minimum Points Using the First or Second Derivative?

Find stationary points by solving $f'(x)=0$, then classify each point using either the first derivative test or the second derivative test. A change from increasing to decreasing gives a local maximum, while decreasing to increasing gives a local minimum.

The Method

First, differentiate $f(x)$ and solve $f'(x)=0$. Substitute each resulting $x$-value into $f(x)$ to obtain the full coordinates of the stationary points.

The first derivative test directly tracks whether the function is increasing or decreasing, so it always classifies a stationary point when the signs are known. The second derivative test is usually faster, because it examines concavity: negative means concave-down and positive means concave-up.

To apply the first derivative test at $x=a$, choose one test value just below $a$ and one just above it, then record the sign of (f'(x)). If the sign does not change, the stationary point is neither a local maximum nor a local minimum.

For the second derivative test, $a$ is a stationary value satisfying $f'(a)=0$. If $f''(a)=0$, the test is , so examine the sign of (f'(x)) on either side instead.

Worked Example

Let

$$f(x)=x^3-3x.$$

Differentiate and find the stationary values:

$$f'(x)=3x^2-3=3(x-1)(x+1).$$

Therefore, $f'(x)=0$ when $x=-1$ or $x=1$. Now calculate

$$f''(x)=6x.$$

At $x=-1$, $f''(-1)=-6<0$, so $(-1$ is a . At , , so is a .

A common misconception is that every solution of $f'(x)=0$ is a maximum or minimum. This is false: for $f(x)=x^3$, , but the function remains increasing through .

IB Exam Technique

This is IB Mathematics AA SL 5.8, applicable to both SL and HL. On Paper 1, show the differentiation, equation $f'(x)=0$, classification test, and coordinates analytically. On Paper 2, a GDC may support the calculation, but examiners still expect the derivative equation and a valid classification using signs or (f''(x)).