Calculus Solver What is an increasing function?
A function is increasing on an interval when larger inputs give larger outputs: if a < b then f(a) < f(b). On a graph it rises left to right. In calculus you find where a function increases by solving f'(x) > 0, because a positive derivative means a positive slope.
Increasing, strictly increasing, non-decreasing
Strictly increasing means a < b always gives f(a) < f(b), with no flat stretches allowed. Non-decreasing, sometimes just called increasing, allows f(a) ≤ f(b), so the function may level off without ever going down.
Courses differ on which word means which, which is a real source of confusion. When it matters, read the inequality rather than the label: < is strict, ≤ is not.
Finding the intervals with the first derivative
The derivative is the slope, so where f'(x) > 0 the function is going up and where f'(x) < 0 it is going down. The method is four steps.
One: differentiate to get f'(x). Two: find the critical points by solving f'(x) = 0 and noting anywhere f'(x) is undefined. Three: those points cut the number line into intervals. Four: test one value inside each interval in f'; a positive result means increasing on that whole interval, negative means decreasing.
You test one point per interval rather than every point because f' can only change sign at a critical point. Between two of them the sign is fixed.
A worked example
Take f(x) = x³ − 3x. Then f'(x) = 3x² − 3 = 3(x − 1)(x + 1), so the critical points are x = −1 and x = 1.
That gives three intervals. Test x = −2: f'(−2) = 9, positive, so increasing. Test x = 0: f'(0) = −3, negative, so decreasing. Test x = 2: f'(2) = 9, positive, so increasing.
The answer is that f increases on (−∞, −1) and on (1, ∞), and decreases on (−1, 1). Write the intervals in terms of x, not y: a common way to lose marks is to give the interval where the outputs live instead of the inputs.
What the endpoints tell you
Where the function switches from increasing to decreasing, it has a local maximum. Where it switches from decreasing to increasing, a local minimum. That is the first derivative test, and it is why finding these intervals is normally the first half of an optimisation problem.
A critical point with the same sign on both sides is neither. f(x) = x³ at x = 0 has f'(0) = 0 but is increasing on both sides, so nothing turns around there.
Reading the sign of the derivative
| f'(x) on an interval | The function is | The graph |
|---|---|---|
| Positive | Increasing | Rising left to right |
| Negative | Decreasing | Falling left to right |
| Zero at a point, sign flips + to − | At a local maximum | A peak |
| Zero at a point, sign flips − to + | At a local minimum | A valley |
| Zero at a point, sign does not flip | Still moving the same way | A flat spot, no turn |
Questions
Can a function be increasing everywhere?
Yes. f(x) = x³ and f(x) = eˣ increase across their whole domain, because their derivatives are never negative. Most functions you meet in a course, though, increase on some intervals and decrease on others.
Should the intervals be open or closed?
Open intervals are the safe answer and the one most courses expect, since the derivative is zero at the endpoints rather than positive. Some textbooks include the endpoints because the function is still non-decreasing there. Follow the convention your course uses.
How is this different from concave up?
Increasing is about the first derivative and means the function is rising. Concave up is about the second derivative and means the rise is getting steeper. A function can rise while flattening out, so the two are independent.
What if the derivative is undefined somewhere?
That point is still a critical point and still splits the number line, so include it. A sharp corner or a vertical tangent can be the place where a function turns around, even though f' does not exist there.
More calculus reference
Stuck on the actual problem
Photograph it and get the worked solution, step by step, with the rule that justified each line named.
