The Reason a Hill Has a Single Highest Point
A ball rolled up a hill speeds toward the top, stops for an instant, then rolls down. That instant where it switches from going up to coming down is where a function turns — and finding those turning points is what increasing and decreasing intervals are all about.
The turning points are where the slope momentarily flattens to zero. Between them, the function is either steadily rising or steadily falling — and the first derivative tells you which, every time.
What Are Increasing and Decreasing Intervals?
An interval is increasing if the output goes up as the input goes up: for every pair $x_1 < x_2$ in the interval, $f(x_1) < f(x_2)$. It's decreasing if the output goes down as the input goes up: $x_1 < x_2$ gives $f(x_1) > f(x_2)$. A function that is increasing (or decreasing) throughout is called monotonic.
The link to calculus is direct. The first derivative $f'(x)$ measures the slope of the curve at each point:
Where $f'(x) > 0$, the slope points uphill, so the function is increasing.
Where $f'(x) < 0$, the slope points downhill, so the function is decreasing.
Where $f'(x) = 0$, the slope is flat — a possible turning point.
Reading slope as direction is the whole idea; everything else is bookkeeping. A function that is strictly increasing or strictly decreasing on its whole domain is automatically a one to one function, because it never revisits an output.
How Do You Find Increasing and Decreasing Intervals?
The first-derivative method is the standard route, and it's the same five steps every time.
Differentiate to get $f'(x)$.
Find the critical points — every $x$ where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only places the function can switch direction.
Split the number line into intervals at those critical points.
Test the sign of $f'(x)$ in each interval by plugging in one convenient point.
Classify: positive $\to$ increasing, negative $\to$ decreasing.
The critical points matter because a continuous function can only change between rising and falling at a point where its slope is zero or undefined. Everywhere else, the direction is locked in. This is the heart of the first derivative test.
What Is a Sign Chart and How Do You Build One?
A sign chart is the bookkeeping tool that turns step 4 into something you can't get wrong. Draw a number line, mark the critical points on it, and in each resulting interval write the sign of $f'(x)$ at a test point.
For $f'(x) = 3(x + 5)(x - 3)$, the critical points are $x = -5$ and $x = 3$. Test one point in each of the three regions:
$x = -6$: $f'(-6) = 3(-1)(-9) = +27 > 0$ — increasing on $(-\infty, -5)$.
$x = 0$: $f'(0) = 3(5)(-3) = -45 < 0$ — decreasing on $(-5, 3)$.
$x = 4$: $f'(4) = 3(9)(1) = +108 > 0$ — increasing on $(3, \infty)$.
The signs flip at each critical point, which confirms both are genuine turning points. You never need the exact value of $f'$ — only its sign.
How Do You Read Intervals Straight From a Graph?
If you already have the graph, skip the calculus: a function is increasing where the curve goes uphill left-to-right and decreasing where it goes downhill. Trace it with your finger moving right; wherever your finger rises, the function increases.
The two methods agree: the spot where the graph turns is exactly the critical point where $f'(x) = 0$. The graph shows you the answer; the derivative proves it.
Should the Intervals Be Open or Closed?
A question that comes up constantly: do you include the critical points in your interval, writing $[-5, 3]$ or $(-5, 3)$? The standard convention is open intervals — exclude the critical points. The cleanest reason: at a turning point the function is momentarily neither increasing nor decreasing (its slope is zero), so it doesn't belong to either label. Writing increasing on $(3, \infty)$ rather than $[3, \infty)$ keeps the categories crisp. Some textbooks include the endpoints when the function is continuous there; both can be defended, but open intervals are the safe default for most courses.
Examples of Increasing and Decreasing Intervals
The examples build from a linear function, through the most common sign-chart mistake, to a cubic and a real-world model.
Example 1
Find the increasing and decreasing intervals of $f(x) = 3x + 5$.
Differentiate: $f'(x) = 3$. The derivative is positive everywhere and never zero, so there are no critical points and no direction changes.
Final answer: increasing on $(-\infty, \infty)$. Every line with positive slope increases everywhere.
Example 2
A common slip — find the intervals for $f(x) = x^2 - 4x + 1$.
Wrong attempt. A student finds $f'(x) = 2x - 4$, sets it to zero to get the critical point $x = 2$, and then writes "increasing on $(-\infty, 2)$, decreasing on $(2, \infty)$" by guessing the order — rising first, falling after, like a hill.
Test it against a sign. At $x = 0$: $f'(0) = 2(0) - 4 = -4 < 0$, which is decreasing, not increasing. The guessed order is backwards.
Correct. Build the sign chart around $x = 2$. At $x = 0$: $f'(0) = -4 < 0$ (decreasing). At $x = 3$: $f'(3) = 2 > 0$ (increasing).
Final answer: decreasing on $(-\infty, 2)$, increasing on $(2, \infty)$. The parabola opens upward, so it falls then rises — the opposite of the hill guess. Always test a point; never assume the order.
Example 3
Find the intervals for $f(x) = x^3 - 3x$.
Differentiate: $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$. Critical points: $x = -1$ and $x = 1$. Sign-test each region:
$x = -2$: $f'(-2) = 3(3)(-1)$... compute as $3(4) - 3 = 9 > 0$ — increasing.
$x = 0$: $f'(0) = -3 < 0$ — decreasing.
$x = 2$: $f'(2) = 9 > 0$ — increasing.
Final answer: increasing on $(-\infty, -1)$ and $(1, \infty)$, decreasing on $(-1, 1)$.
Example 4
Find the intervals for $f(x) = x^3 + 3x^2 - 45x + 9$.
Differentiate: $f'(x) = 3x^2 + 6x - 45 = 3(x + 5)(x - 3)$. Critical points: $x = -5$ and $x = 3$. The sign chart built in the section above applies directly.
Final answer: increasing on $(-\infty, -5)$ and $(3, \infty)$, decreasing on $(-5, 3)$.
Example 5
Where is $f(x) = e^x$ increasing or decreasing?
Differentiate: $f'(x) = e^x$. Since $e^x > 0$ for every real $x$, the derivative is never zero or negative.
Final answer: increasing on $(-\infty, \infty)$, with no decreasing interval. Exponential growth never turns around — which is exactly why it's also one-to-one.
Example 6
A company's profit (in thousands) is $P(t) = -t^2 + 8t - 5$, where $t$ is months since launch. When is profit rising, and when does it start to fall?
Differentiate: $P'(t) = -2t + 8$. Set to zero: $t = 4$. Test signs — at $t = 0$, $P'(0) = 8 > 0$ (rising); at $t = 6$, $P'(6) = -4 < 0$ (falling).
Final answer: profit increases on $(0, 4)$ and decreases on $(4, \infty)$ — it peaks at month $4$. The turning point is the month to watch, which is the kind of question businesses ask of every revenue model.
Where Increasing and Decreasing Intervals Earn Their Place
"When does this stop growing and start to fall?"
The first derivative test grew out of the calculus developed independently by Isaac Newton (1643–1727, England) and Gottfried Wilhelm Leibniz (1646–1716, Germany) in the late 17th century. Both realized the same thing: the slope of a curve, captured by the derivative, tells you instantly whether a quantity is rising or falling — without computing a single value of the function itself.
Where the idea does real work:
Optimization everywhere. Maximum profit, minimum cost, peak dosage, shortest path — every optimization problem comes down to finding where a quantity stops increasing and starts decreasing. The turning point is the answer.
Economics and business. Marginal analysis — "is the next unit adding or subtracting value?" — is reading the sign of a derivative. Example 6 is the everyday version.
Physics and motion. Velocity is the derivative of position; where velocity changes sign, an object reverses direction. The ball on the hill is literally this.
Where Students Trip Up on Increasing and Decreasing Intervals
Mistake 1: Guessing the order instead of testing a sign
Where it slips in: After finding the critical point, deciding which side is increasing by intuition rather than a sign test.
Don't do this: Assume the function rises before the critical point and falls after (the "hill" picture), without checking.
The correct way: Always plug a test point into $f'(x)$ for each interval and read the sign. An upward parabola falls then rises — the opposite of the hill (Example 2).
Mistake 2: Forgetting critical points where the derivative is undefined
Where it slips in: Functions with a sharp corner or a vertical tangent — like $f(x) = x^{2/3}$ — where $f'(x)$ is undefined at a point but never equals zero.
Don't do this: Only solve $f'(x) = 0$ and ignore where $f'(x)$ doesn't exist.
The correct way: Critical points are where $f'(x) = 0$ or $f'(x)$ is undefined. The second-guesser who solves only the equation misses turning points hiding at corners.
Mistake 3: Including critical points inside the intervals
Where it slips in: Writing the answer as $[-5, 3]$ instead of $(-5, 3)$.
Don't do this: Reflexively close every interval at the critical points.
The correct way: Use open intervals by default — at a turning point the slope is zero, so the function is momentarily neither rising nor falling.
Key Takeaways
A function's increasing and decreasing intervals are where its output rises or falls, found by the sign of the first derivative $f'(x)$.
The method is five steps: differentiate, find critical points, split the number line, test a sign in each interval, classify.
Critical points occur where $f'(x) = 0$ or is undefined — the only places direction can change.
The most common mistake is guessing the order instead of testing a sign; an upward parabola falls before it rises.
Intervals are written open by default, and the whole idea powers optimization, marginal analysis, and motion problems.
Practice These Before Moving On
Find the increasing and decreasing intervals of $f(x) = x^2 - 6x + 2$.
Find the intervals for $f(x) = -x^3 + 3x$.
State where $f(x) = \ln x$ (for $x > 0$) is increasing or decreasing.
Answer to Question 1: decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$. Answer to Question 2: $f'(x) = -3x^2 + 3 = -3(x-1)(x+1)$, so decreasing on $(-\infty, -1)$ and $(1, \infty)$, increasing on $(-1, 1)$. Answer to Question 3: $f'(x) = 1/x > 0$ for all $x > 0$, so increasing on $(0, \infty)$. If Question 1 came out backwards, return to Mistake 1 and test a point.
Want a live Bhanzu trainer to walk through more increasing and decreasing intervals problems? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content