Proving a function is decreasing comes down to showing that as your input gets larger, the output gets smaller. The most common method uses calculus: if the first derivative is negative on an interval, the function is decreasing there. But depending on the context, you might use the formal definition directly, a sign chart, or even mathematical induction for sequences. Here’s how each approach works.
The Formal Definition
A function is decreasing on an interval when larger inputs always produce smaller (or equal) outputs. Written precisely: for all x₁ and x₂ in the domain where x₁ < x₂, we have f(x₁) ≥ f(x₂). If the inequality is strict, meaning f(x₁) > f(x₂) whenever x₁ < x₂, the function is called strictly decreasing.
This distinction matters. A strictly decreasing function always goes down. A non-increasing function (the looser version) can have flat sections where consecutive values are equal, as long as it never goes up. In most calculus courses, “decreasing” means the strict version unless stated otherwise.
You can use this definition directly as a proof method. Pick two arbitrary points x₁ < x₂ in the interval, then algebraically show that f(x₁) > f(x₂). This approach works well for simple functions but gets tedious for complicated ones, which is where calculus comes in.
Using the First Derivative
The most widely used method relies on a clean rule: if f'(x) < 0 for every x in an open interval, then f is decreasing on that interval. The logic behind this connects to the Mean Value Theorem. For any two points x₁ < x₂ in the interval, the theorem guarantees some point c between them where:
f'(c) = (f(x₂) – f(x₁)) / (x₂ – x₁)
Since x₂ – x₁ is positive and f'(c) is negative, the numerator f(x₂) – f(x₁) must be negative. That means f(x₂) < f(x₁), which is exactly the definition of decreasing.
So in practice, if you can show the derivative is negative everywhere on an interval, you’re done. For example, f(x) = 1/x has f'(x) = -1/x², which is negative for all x ≠ 0. By the theorem, f is decreasing on (-∞, 0) and separately on (0, ∞).
Step-by-Step: Finding Decreasing Intervals
Most textbook problems ask you to find where a function is decreasing, not just confirm that it is. Here’s the standard process.
Step 1: Find the derivative. Differentiate f(x) using whatever rules apply (power rule, chain rule, quotient rule, etc.).
Step 2: Find critical numbers. Set f'(x) = 0 and solve. Also note any x-values where f'(x) is undefined. These critical numbers are the only places where the function can switch between increasing and decreasing.
Step 3: Factor the derivative. If f'(x) is a polynomial, factor it completely. If it’s a fraction, factor the numerator and denominator separately. This makes the next step much easier.
Step 4: Build a sign chart. Draw a number line and mark each critical number. This splits the domain into intervals. In each interval, pick any convenient test value, plug it into f'(x), and check whether the result is positive or negative.
Step 5: Read the results. Every interval where f'(test value) < 0 is an interval where f is decreasing.
A Worked Example
Take f(x) = x³ – 3x². The derivative is f'(x) = 3x² – 6x, which factors as 3x(x – 2). Setting f'(x) = 0 gives critical numbers at x = 0 and x = 2. These split the real line into three intervals: (-∞, 0), (0, 2), and (2, ∞).
Now test each interval. Pick x = -1 for the first: f'(-1) = 3(-1)(-1 – 2) = 3(-1)(-3) = 9, which is positive, so f is increasing on (-∞, 0). Pick x = 1 for the second: f'(1) = 3(1)(1 – 2) = -3, which is negative, so f is decreasing on (0, 2). Pick x = 3 for the third: f'(3) = 3(3)(3 – 2) = 9, positive again, so f is increasing on (2, ∞).
The function is decreasing only on the interval (0, 2).
Proving a Sequence Is Decreasing
For discrete sequences rather than continuous functions, you have two main tools: direct comparison of consecutive terms, and mathematical induction.
The direct approach is straightforward. Show that aₙ₊₁ < aₙ for all valid n. You can do this by computing aₙ₊₁ – aₙ and showing the result is always negative, or by computing aₙ₊₁/aₙ and showing it’s always less than 1 (when all terms are positive).
For sequences defined by a recursive formula, induction often works better. Start with the base case: verify that a₂ ≤ a₁. Then assume the inequality holds at step k, meaning aₖ ≥ aₖ₊₁, and use the recursive definition to show that aₖ₊₁ ≥ aₖ₊₂. Once you’ve shown both pieces, the principle of induction guarantees the sequence is decreasing for all n.
A third option applies when the sequence is defined by a formula like aₙ = f(n). You can treat f as a continuous function, take its derivative, and show f'(x) < 0 for all x ≥ 1. If the continuous version is decreasing, the sequence of its values at integers is too.
What the Second Derivative Tells You
The first derivative tells you whether a function is decreasing. The second derivative tells you how it’s decreasing. Specifically, it reveals whether the rate of decrease is speeding up or slowing down.
If f'(x) < 0 and f”(x) < 0 on an interval, the function is decreasing and concave down. Picture the left side of an upside-down bowl: the curve drops and keeps getting steeper. The rate of decrease is accelerating.
If f'(x) < 0 and f”(x) > 0, the function is decreasing but concave up. This looks like the right side of a regular bowl: the curve is still dropping, but it’s flattening out and approaching a bottom. The rate of decrease is slowing down.
This matters when you need to describe the behavior of a function in more detail, or when a problem asks whether a decreasing function is “decreasing at an increasing rate” versus “decreasing at a decreasing rate.” The second derivative answers that question.
Common Pitfalls
A function can be decreasing on two separate intervals without being decreasing across both. The function 1/x is decreasing on (-∞, 0) and decreasing on (0, ∞), but it’s not decreasing on its entire domain, because f(-1) = -1 is less than f(1) = 1. The jump at the discontinuity breaks the chain.
Another common mistake is confusing “the derivative is zero at a point” with “the function stops decreasing.” A derivative that equals zero at isolated points (while being negative everywhere else on the interval) doesn’t interrupt the decreasing behavior. The function f(x) = -x³ has f'(x) = -3x², which equals zero at x = 0 but is negative everywhere else. The function is still decreasing on all of ℝ.
Finally, remember that the derivative method requires the function to be differentiable on the interval. For piecewise functions or functions with corners, you may need to fall back on the formal definition and check the inequality f(x₁) > f(x₂) directly at the trouble spots.