How Do You Find the Concavity of a Function? Step-by-Step Guide

Concavity tells you which way a curve bends. To find it, take the second derivative of the function, set it equal to zero, find the points where it changes sign, then test the sign of the second derivative on each interval between those points. If the second derivative is positive on an interval, the function is concave up (bending like a cup); if it’s negative, the function is concave down (bending like a frown). That sign test is the whole method, and the rest of this guide walks through it step by step.

The Core Rule: The Second Derivative Test for Concavity

The second derivative measures how the slope itself is changing. When $f’‘(x) > 0$, the slope is increasing, so the curve turns upward. When $f’'(x) < 0$, the slope is decreasing, so the curve turns downward. This is a standard result covered in most calculus courses, and Khan Academy treats it as the foundation for the concavity unit.

A quick way to remember the shapes:

• Concave up ($f’’ > 0$): the curve holds water, like the letter U.
• Concave down ($f’’ < 0$): the curve spills water, like an upside-down U.

The points where concavity actually switches are called inflection points, and they’re the natural landmarks you’re hunting for.

Step-by-Step: How Do You Find the Concavity of a Function?

Here is the full procedure, in order.

Step 1: Find the second derivative

Differentiate the function once to get $f’(x)$, then differentiate again to get $f’‘(x)$. If the function is a polynomial, this is usually quick. For example, if $f(x) = x^3 - 3x$, then $f’(x) = 3x^2 - 3$ and $f’'(x) = 6x$.

Step 2: Set the second derivative to zero and solve

Solve $f’'(x) = 0$ for all real values of $x$. These are your candidate inflection points. In the example, $6x = 0$ gives $x = 0$.

Step 3: Check where the second derivative is undefined

The second derivative can also change sign where it fails to exist, such as where a denominator hits zero. Those points belong on your list of candidates too. Be careful here: a point where $f’'$ is undefined isn’t automatically an inflection point, but it must be tested.

Step 4: Divide the number line into intervals

Take all the candidate points and use them to split the domain into open intervals. In the example, the single candidate $x = 0$ splits the number line into two intervals: $(-\infty, 0)$ and $(0, \infty)$.

Step 5: Test the sign on each interval

Pick any convenient test value inside each interval and plug it into $f’‘(x)$. You only care about the sign, not the exact number. For $x < 0$, pick $x = -1$: $f’‘(-1) = -6$, which is negative, so the function is concave down there. For $x > 0$, pick $x = 1$: $f’'(1) = 6$, which is positive, so the function is concave up there.

Step 6: Identify the inflection points

If the sign of $f’'$ changes as you cross a candidate point, and the function is continuous there, that point is an inflection point. In the example, the sign flips from negative to positive at $x = 0$, so $(0, 0)$ is an inflection point. The function $f(x) = x^3 - 3x$ is concave down on $(-\infty, 0)$ and concave up on $(0, \infty)$.

That’s the entire workflow. The MIT OpenCourseWare calculus materials frame it the same way: locate the sign changes of the second derivative, then label the intervals.

Common Mistakes and a Few Warnings

A few traps trip people up repeatedly.

• Stopping at $f’'(x) = 0$ without testing. A zero of the second derivative is only a candidate. If the sign doesn’t change there, it’s not an inflection point. For instance, $f(x) = x^4$ has $f’'(x) = 12x^2$, which equals zero at $x = 0$ but stays positive on both sides, so $x^4$ is concave up everywhere and has no inflection point.
• Confusing concavity with increasing or decreasing. Those come from the first derivative. Concavity comes from the second. A function can be increasing while concave down, or decreasing while concave up.
• Forgetting undefined points. Rational and radical functions can change concavity at points that never appear when you set $f’'(x) = 0$.

If you want a visual confirmation, graphing the second derivative alongside the original function makes the correspondence obvious: where $f’'$ dips below the $x$-axis, the original curve bends downward. Wolfram MathWorld defines the inflection point formally as a point where the curve changes from concave up to concave down or vice versa, which matches the sign-change test above.

Key Takeaways

• Concavity is determined entirely by the sign of the second derivative, $f’'(x)$.
• $f’‘(x) > 0$ means concave up; $f’'(x) < 0$ means concave down.
• The method: find $f’‘(x)$, solve $f’‘(x) = 0$, note where $f’'$ is undefined, split the domain into intervals, and test the sign in each one.
• An inflection point occurs only where the sign of $f’'$ actually changes and the function is continuous.
• A zero of the second derivative without a sign change is not an inflection point — always verify with a test value.