Do I use open or closed brackets for the intervals?

Standard practice usually uses open parentheses, like $(0, 2)$, because at the exact critical point, the slope is zero (neither increasing nor decreasing). However, some textbook authors use closed brackets to indicate the function includes the endpoints. Check your specific class syllabus to see what your teacher prefers!

What if the derivative is never zero?

If the derivative is never zero and never undefined, it means the function never changes direction. It is either always increasing or always decreasing across its entire domain. A simple straight line like $f(x) = 2x$ is a good example; its derivative is always $2$, so it is always increasing.

Can a critical point NOT be a change in direction?

Yes, absolutely! If you take the function $f(x) = x^3$, the derivative is $f'(x) = 3x^2$. Setting it to zero gives a critical point at $x = 0$. However, if you test $-1$ and $1$, the derivative is positive for both. The graph flattens out for a single moment and then keeps going up. This is why testing every interval is mandatory.

How do you find where a function is increasing or decreasing?

To find where a function is increasing or decreasing, you look at its first derivative. If the first derivative is positive, the function is going up; if the derivative is negative, the function is going down.

Think of the derivative as the speedometer of a car. If your velocity is positive, you are moving forward, so your position is increasing. If your velocity is negative, you are moving in reverse, so your position is decreasing. The exact moments where you switch from one to the other are the moments your speed hits zero.

What it means mathematically

In calculus, the derivative $f'(x)$ represents the slope of the tangent line to the original function $f(x)$ . A positive slope means the graph is pointing upward as you move from left to right. A negative slope means it is pointing downward. Therefore, a function is increasing on any interval where $f'(x) > 0$ and decreasing on any interval where $f'(x) < 0$ .

Finding the critical points

Before you can find where the function is positive or negative, you need to find where it could possibly switch between the two. These switching places are called critical points. You find them by taking the derivative $f'(x)$ and setting it equal to zero. You must also look for places where $f'(x)$ is undefined, such as a denominator equaling zero, because the function's direction can jump or change there too.

Testing the intervals

Once you have your critical points, you place them on a number line. This chops the number line into separate sections or intervals. Because the derivative can only change signs at a critical point, it will be consistently positive or consistently negative across an entire interval. You just pick one test number from inside each interval, plug it into the derivative $f'(x)$ , and see if the result is positive or negative.

Where students slip

A common mistake is plugging the test numbers back into the original function $f(x)$ instead of the derivative $f'(x)$ . Remember, the original function tells you the height of the graph, but the derivative tells you the direction. Always test your numbers in $f'(x)$ . Another trap is assuming the signs will always alternate (positive, then negative, then positive). They often do, but not always, so you must test every single interval manually.

Worked through

Find the intervals where the function $f(x) = x^3 - 3x^2 + 1$ is increasing and decreasing.

First, we find the derivative: $f'(x) = 3x^2 - 6x$ .

Next, we find the critical points by setting the derivative to zero: $3x^2 - 6x = 0$ $3x(x - 2) = 0$

This gives us critical points at $x = 0$ and $x = 2$ .

Now, we set up our intervals on a number line: $(-\infty, 0)$ , $(0, 2)$ , and $(2, \infty)$ . We choose a test point in each interval and plug it into $f'(x)$ .

Test $x = -1$ for the first interval: $f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9$ . Since $9 > 0$ , the function is increasing on $(-\infty, 0)$ .

Test $x = 1$ for the second interval: $f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$ . Since $-3 < 0$ , the function is decreasing on $(0, 2)$ .

Test $x = 3$ for the third interval: $f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9$ . Since $9 > 0$ , the function is increasing on $(2, \infty)$ .

Final Answer: The function is increasing on $(-\infty, 0)$ and $(2, \infty)$ , and decreasing on $(0, 2)$ .

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Where this comes from: OpenStax Calculus Volume 1, Chapter 4: Applications of Derivatives · Khan Academy, Unit: Applications of derivatives