What Are Increasing And Decreasing Functions?
Increasing and decreasing functions describe the direction a function travels as the input grows. A function $f$ is increasing on an interval if larger inputs give larger outputs: for any $x_1 < x_2$ in the interval, $f(x_1) < f(x_2)$. It is decreasing on an interval if larger inputs give smaller outputs: for any $x_1 < x_2$, $f(x_1) > f(x_2)$.
For a differentiable function, calculus turns those word definitions into a single test on the slope. The derivative $f'(x)$ is the slope of the tangent line, so its sign reports the direction of travel:
$$f'(x) > 0 \ \text{on } (a,b) \ \Rightarrow\ f \text{ is increasing on } (a,b)$$
$$f'(x) < 0 \ \text{on } (a,b) \ \Rightarrow\ f \text{ is decreasing on } (a,b)$$
The geometric picture sits right beside the algebra: a positive slope points the tangent uphill, a negative slope points it downhill, and a zero slope marks the flat spot where the curve may turn. This sign-of-the-slope idea is the whole engine of the topic. It is the local, interval-by-interval view of a monotonic function, and it powers the first derivative test for locating peaks and valleys.
How Do You Find Increasing And Decreasing Intervals From f'(x)?
The procedure is the same every time. It rests on one fact: a differentiable function can only switch between rising and falling where its derivative is zero or undefined, so those points are the only possible boundaries.
Differentiate. Compute $f'(x)$. If you need the mechanics, the derivative reference collects the rules.
Find the critical points. Solve $f'(x) = 0$, and separately note any $x$ where $f'(x)$ is undefined but $f$ is defined. These are the critical points, the candidate boundaries.
Split the domain. The critical points (together with any gaps in the domain) cut the number line into open subintervals.
Test the sign on each subinterval. Pick one convenient test value inside each subinterval and evaluate the sign of $f'$ there. One point speaks for the whole subinterval, because $f'$ cannot change sign without passing through a critical point.
Read the direction. Positive sign means increasing on that subinterval; negative means decreasing.
The output is written as intervals, never as single points. The standard convention uses open intervals, since at a turning point the slope is zero and the curve is momentarily neither rising nor falling.
Worked Examples: Finding The Increasing And Decreasing Intervals
Each example follows the five steps and states the intervals as the final answer.
Example 1: A cubic with two turning points.
Find where $f(x) = x^3 - 3x$ is increasing and decreasing.
Differentiate and factor:
$$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$
Setting $f'(x) = 0$ gives critical points $x = -1$ and $x = 1$. These split the line into three subintervals. Test one value in each:
$$f'(-2) = 3(-3)(-1) = 9 > 0, \quad f'(0) = 3(-1)(1) = -3 < 0, \quad f'(2) = 3(1)(3) = 9 > 0$$
The signs read $+, -, +$. So $f$ increases, turns at $x=-1$, decreases, turns at $x=1$, then increases again.
Final answer: increasing on $(-\infty, -1)$ and $(1, \infty)$; decreasing on $(-1, 1)$. The turning points are a local maximum at $(-1, 2)$ and a local minimum at $(1, -2)$.
Example 2: A product where one factor is always positive.
Find where $f(x) = x e^{-x}$ is increasing and decreasing.
Differentiate with the product rule:
$$f'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x)$$
The factor $e^{-x}$ is positive for every real $x$, so the sign of $f'$ is decided entirely by $(1 - x)$. Setting $f'(x) = 0$ gives the single critical point $x = 1$. Test each side:
$$f'(0) = e^{0}(1) = 1 > 0, \qquad f'(2) = e^{-2}(-1) < 0$$
Final answer: increasing on $(-\infty, 1)$, decreasing on $(1, \infty)$, with a local maximum at $x = 1$ of height $f(1) = e^{-1} \approx 0.3679$.
Example 3: A boundary where the derivative is undefined.
Find where $f(x) = x + \dfrac{4}{x}$ is increasing and decreasing.
The domain excludes $x = 0$. Differentiate:
$$f'(x) = 1 - \frac{4}{x^2}$$
Solve $f'(x) = 0$: this gives $x^2 = 4$, so $x = -2$ and $x = 2$. The derivative is also undefined at $x = 0$, which is already a gap in the domain, so it too is a boundary. Four subintervals result. Test each:
$$f'(-3) = 1 - \tfrac{4}{9} > 0, \quad f'(-1) = 1 - 4 = -3 < 0, \quad f'(1) = 1 - 4 = -3 < 0, \quad f'(3) = 1 - \tfrac{4}{9} > 0$$
Final answer: increasing on $(-\infty, -2)$ and $(2, \infty)$; decreasing on $(-2, 0)$ and $(0, 2)$. Note that $x = 0$ is a boundary even though $f'(0)$ never equals zero: it is excluded from the domain, and skipping it would merge two subintervals that behave differently. The local maximum is $(-2, -4)$ and the local minimum is $(2, 4)$.
What Is The Sign Test For Increasing And Decreasing Functions?
The table below is the whole method on one card. It also captures the case competitors skip: where $f'$ is undefined.
Table: Reading the direction of a function from the sign of its first derivative.
Sign of $f'(x)$ on an interval | Slope of the tangent | Behaviour of $f$ |
|---|---|---|
$f'(x) > 0$ | uphill (positive) | increasing |
$f'(x) < 0$ | downhill (negative) | decreasing |
$f'(x) = 0$ at a point | flat (horizontal tangent) | critical point, possible turn |
$f'(x)$ undefined at a point (with $f$ defined) | corner, cusp, or vertical tangent | critical point, possible turn |
Read the sign column first. The direction of the function is nothing more than the sign of its slope, held steady across a whole subinterval. This same sign chart feeds directly into maxima and minima and into full curve sketching, where the pattern of rises and falls builds the shape of the graph.
Why Does The Sign Of The Derivative Decide The Direction?
It is tempting to treat "positive slope means increasing" as a definition, but it is a theorem, and the reason it holds is worth seeing.
The Mean Value Theorem is the engine. Take any two points $x_1 < x_2$ in an interval where $f' > 0$. The Mean Value Theorem guarantees a point $c$ between them with $f(x_2) - f(x_1) = f'(c),(x_2 - x_1)$. Since $f'(c) > 0$ and $x_2 - x_1 > 0$, the right side is positive, so $f(x_2) > f(x_1)$. That is exactly the definition of increasing, now proven rather than assumed.
Geometry says the same thing. The tangent line is the curve's best local straight-line match. If every tangent on an interval leans uphill, the curve cannot secretly drop, so it rises across the whole interval.
One test value is enough because of continuity. When $f'$ is continuous, it can only change sign by passing through zero. Between two consecutive critical points it never hits zero, so it keeps one sign, and a single test value reveals it.
The link runs both directions of calculus at once: the derivative is built from slopes, and the sign of that slope reconstructs the rise and fall of the original function. This is one reason increasing and decreasing behaviour sits at the centre of applications of derivatives.
Who Discovered How To Find A Function's Rise And Fall?
Long before derivatives had their modern name, mathematicians were hunting for the flat spots where a curve turns from rising to falling.
Two later figures made the sign test rigorous:
Michel Rolle (1652–1719, France) proved that between two points with equal function values a differentiable curve must have a horizontal tangent, the result now called Rolle's theorem.
Joseph-Louis Lagrange (1736–1813, Italy and France) generalised it into the Mean Value Theorem, which is the exact statement that turns "$f' > 0$" into "$f$ is increasing," as shown in the section above.
Where Are Increasing And Decreasing Functions Used In The Real World?
Anywhere a quantity rises and falls over time or input, the sign of a derivative is doing the classification.
Economics: a firm's profit is increasing while marginal profit is positive and starts to fall once it turns negative, so the profit-maximising output is exactly the turning point where the derivative changes sign.
Physics and motion: position is increasing while velocity (its derivative) is positive, and an object reverses direction at the instant velocity passes through zero.
Medicine and growth: an infection or population curve is increasing while its growth rate is positive; the peak, where the rate hits zero, is what public-health models race to predict.
Engineering optimization: a design is improving while the derivative of its performance measure is positive, so engineers push a parameter until that sign flips.
Business analytics: whether revenue, traffic, or temperature is climbing or dropping on any interval is read straight from the sign of its rate of change.
One sign test, applied to a rate of change, tells every field whether the thing it cares about is currently getting bigger or smaller.
What Are The Most Common Mistakes With Increasing And Decreasing Functions?
These four errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and course review guides.
Assuming every critical point is a maximum or minimum.
Where it slips in:
A student finds $f'(x) = 0$ and immediately labels the point a peak or a valley.
Don't do this:
Do not conclude a turn from $f'(x) = 0$ alone. A zero slope is only a candidate for a turn.
The correct way:
Check whether $f'$ actually changes sign there. For $f(x) = x^3$, $f'(x) = 3x^2$ gives $f'(0) = 0$, but $f'$ is positive on both sides of $0$, so the function keeps increasing through the origin with no extremum at all.
Saying a function is "increasing at a point."
Where it slips in:
A student writes "$f$ is increasing at $x = 2$" as if a single point had a direction.
Don't do this:
Do not attach increasing or decreasing to one $x$-value. Direction is a property of an interval, not a point.
The correct way:
State the behaviour on an open interval, for example "increasing on $(1, \infty)$." A point only has a slope; an interval has a direction.
Guessing the order instead of testing the signs.
Where it slips in:
After finding the critical points, a student assumes the curve rises first and falls after, picturing a single hill.
Don't do this:
Do not assume the up-then-down pattern. Many functions fall first, or alternate several times.
The correct way:
Test one value in every subinterval and let the signs of $f'$ dictate the order. In Example 1 the pattern was $+, -, +$, not a single hill.
Checking only where $f'(x) = 0$ and ignoring where $f'$ is undefined.
Where it slips in:
A student solves $f'(x) = 0$, splits the line at those roots only, and forgets points where $f'$ blows up or the domain has a gap.
Don't do this:
Do not treat $f'(x) = 0$ as the only source of boundaries. A corner, a cusp, or a domain gap is a boundary too.
The correct way:
Include every $x$ where $f'$ is undefined while $f$ is defined, plus the domain's own breaks. In Example 3, $x = 0$ had to be a boundary even though $f'$ is never zero there.
Practice Problems On Increasing And Decreasing Functions
Work each one, then check against the answer. Answers are verified.
Find the intervals for $f(x) = x^2 - 6x + 5$.
(Answer: $f'(x) = 2x - 6$, critical point $x = 3$; decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$.)Find the intervals for $f(x) = 2x^3 - 9x^2 + 12x$.
(Answer: $f'(x) = 6(x-1)(x-2)$, critical points $x = 1, 2$; increasing on $(-\infty, 1)$, decreasing on $(1, 2)$, increasing on $(2, \infty)$.)Find the intervals for $f(x) = x^3 - 12x$.
(Answer: $f'(x) = 3(x-2)(x+2)$, critical points $x = \pm 2$; increasing on $(-\infty, -2)$, decreasing on $(-2, 2)$, increasing on $(2, \infty)$.)Find the intervals for $f(x) = e^{x} - x$.
(Answer: $f'(x) = e^{x} - 1$, critical point $x = 0$; decreasing on $(-\infty, 0)$, increasing on $(0, \infty)$.)Where is $f(x) = \ln x - x$ increasing?
(Answer: domain $x > 0$; $f'(x) = \tfrac{1}{x} - 1$, critical point $x = 1$; increasing on $(0, 1)$, decreasing on $(1, \infty)$.)
Where Should You Go Next After Increasing And Decreasing Functions?
The sign test is a gateway to the rest of curve analysis, and several natural doors open from here.
First derivative test. Use the same sign changes to classify each critical point as a maximum, a minimum, or neither.
Maxima and minima. Turn the turning points into the highest and lowest values a function reaches.
Stationary points and curve sketching. Combine rise-and-fall information with concavity to draw a full graph by hand.
If your child is meeting increasing and decreasing functions for the first time, a live Bhanzu trainer teaches the sign chart from the slope idea up, so the method feels like reading a graph rather than following a recipe, in the Bhanzu math program.
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