How To Prove A Function Is Decreasing Using Derivatives?

To prove a function is decreasing using derivatives, you show that its derivative is negative on the interval in question. If f′(x) < 0 for every x in an open interval, then f is decreasing on that interval. The derivative measures the function’s rate of change, so a negative rate at every point means the output values are falling as the input values rise.

That is the whole test. The work is in finding the derivative, finding where it is negative, and being careful about the boundaries of the interval. This article walks through each step, the exact definitions, and the traps that cause students to lose points on exams.

What Does It Actually Mean For A Function To Be Decreasing?

A function is decreasing on an interval when larger inputs always produce smaller outputs within that interval. Formally, for any two points a and b in the interval with a < b, you have f(a) > f(b).

This is a strict definition, and the strictness matters. Some textbooks define “decreasing” loosely to allow flat spots, calling it non-increasing, and reserve “strictly decreasing” for the version above. Others use “decreasing” to mean strictly decreasing. Before you write an answer on an exam, check which convention your course uses. The distinction changes whether a flat stretch counts as decreasing.

There is also a difference between decreasing on an interval and decreasing everywhere. A function can fall on one stretch, rise on another, and fall again later. Calculus questions almost always ask about a specific interval, so identify that interval first.

How Does The Derivative Test For Decreasing Functions Work?

The derivative test rests on a theorem you can state plainly: if f′(x) < 0 for all x in an open interval, then f is decreasing on that interval. This is one of the standard results connecting derivatives to function behavior, and it appears in every first-year calculus text.

The reasoning is intuitive once you see it. The derivative is the limit of the slope of secant lines as the gap between two points shrinks. If that limit is negative at every point, the tangent line tilts downward everywhere, and the curve must be heading down as you move right. The formal proof uses the Mean Value Theorem, which guarantees that between any two points there is a spot where the instantaneous rate of change equals the average rate of change over that span. If the instantaneous rate is always negative, the average rate must be too, which forces f(a) > f(b).

One clarification that trips people up: the test uses a strict inequality. If f′(x) < 0 on an open interval, the function is strictly decreasing there. If the derivative is only non-positive, allowing zeros, you can only conclude the function is non-increasing. A single point where the derivative equals zero does not break the decreasing behavior on the surrounding interval, but a whole stretch of zeros does.

How To Prove A Function Is Decreasing Using Derivatives: Step By Step

The proof follows a repeatable sequence. Work through these steps in order and you will not miss anything.

• Find the derivative. Differentiate the function using the standard rules — power, product, quotient, chain, or whatever applies.
• Set the derivative equal to zero. Solve f′(x) = 0 to find critical points. These are the only places where the derivative can change sign.
• Note where the derivative is undefined. Division by zero, square roots of negatives, and logarithms of non-positive numbers all create points where the derivative does not exist. These also mark possible sign changes.
• Build a sign chart. The critical points and undefined points divide the number line into intervals. Pick one test value inside each interval and evaluate the derivative there.
• Conclude. Every interval where the test value gives a negative derivative is an interval where the function is decreasing.

Pick test values that are easy to compute. If your critical points are 2 and 5, test x = 0, x = 3, and x = 6. You only need the sign of the result, not its exact value, so round aggressively if it saves time.

Can You Prove A Function Is Decreasing Without A Sign Chart?

Yes, and sometimes it is cleaner. If you can show algebraically that the derivative is negative for all x in the interval, you have your proof without testing sample points.

Consider f(x) = −x³ on the interval (0, ∞). The derivative is f′(x) = −3x². Since x² is always positive for nonzero x, −3x² is always negative. No sign chart needed — the algebra does the work.

This approach works best when the derivative has an obvious sign. A negative constant times a squared term, a negative number divided by a positive quantity, or any expression you can bound below zero all qualify. When the derivative is a complicated polynomial or a rational function with several factors, the sign chart is usually faster and less error-prone.

What Common Mistakes Do Students Make With This Test?

The most frequent error is testing the wrong thing. Students sometimes plug values into the original function instead of the derivative. The original function tells you where the curve is above or below the x-axis. The derivative tells you where it is rising or falling. These are completely different questions.

Another common slip is forgetting that the conclusion applies to intervals, not single points. It is meaningless to say a function is decreasing at x = 3. It is decreasing on an interval that contains 3.

A third mistake involves endpoints. The derivative test applies to open intervals. If a question asks about a closed interval like [1, 5], you determine decreasing behavior on the open interval (1, 5) and then check whether the endpoints extend that behavior. Many instructors accept the open-interval answer for the closed-interval question, but not all do.

Finally, watch for functions where the derivative is negative on one interval and positive on the next, with a critical point in between. The function is decreasing up to that point and increasing after it. A local minimum sits at the transition. Stating that the function is “decreasing” without naming the interval is incomplete.

How Do You Handle Functions With No Derivative At Some Points?

Some functions have corners, cusps, or vertical tangents where the derivative does not exist. The absolute value function is the classic example. At x = 0, the graph has a sharp corner, and the derivative is undefined there.

The derivative test still works, but you must treat these points as boundaries. Split the domain at every point where the derivative fails to exist, then run the sign test on each resulting piece. For f(x) = |x|, the derivative is −1 for x < 0 and +1 for x > 0. The function is decreasing on (−∞, 0) and increasing on (0, ∞). The point x = 0 belongs to neither interval.

This matters because a function can be decreasing on both sides of a point where the derivative does not exist, or decreasing on one side and increasing on the other. The sign chart handles both cases cleanly as long as you include the undefined points as dividers.

Does A Negative Derivative Guarantee The Function Never Rises?

On the interval where the derivative is negative, yes. The theorem is an if-and-only-if in one direction: negative derivative on an open interval guarantees decreasing behavior on that interval.

The converse is where students get confused. A function can be decreasing on an interval even if the derivative is zero at isolated points inside it. For example, f(x) = −x³ has derivative −3x², which equals zero at x = 0 but is negative everywhere else. The function is strictly decreasing on the entire real line, including through x = 0.

What you cannot have is a function that is strictly decreasing on an interval while its derivative is positive on a subinterval of positive length. That would contradict the theorem. If the derivative is positive on a stretch, the function rises there.

This is why the sign chart approach is reliable. It tells you exactly where the derivative is negative, and those are exactly the intervals where the function decreases. No guesswork required.