Increasing and decreasing functions
If f′(x) > 0 on an interval, f is strictly increasing there. If f′(x) < 0, f is strictly decreasing.
What the words mean
Let I be an interval in the domain of a real function f. Read its graph from left to right. If the graph rises, f is increasing on I. If it falls, f is decreasing. The word strictly means the graph is never flat: a larger x always gives a larger value (or always a smaller one). A function can increase on one interval and decrease on another, so always say on which interval.
The four definitions
| Name | For every x₁ < x₂ in I | Examples on R |
|---|---|---|
| Increasing | f(x₁) ≤ f(x₂) | Every strictly increasing function; also a constant such as f(x) = 7 |
| Strictly increasing | f(x₁) < f(x₂) | f(x) = 2x + 3, f(x) = x³, f(x) = eˣ |
| Decreasing | f(x₁) ≥ f(x₂) | Every strictly decreasing function; also f(x) = 7 |
| Strictly decreasing | f(x₁) > f(x₂) | f(x) = 1 − 3x, f(x) = e⁻ˣ |
The derivative test
- Let f be continuous on [a, b] and differentiable on (a, b).
- If f′(x) > 0 for every x in (a, b), f is increasing on [a, b]. The proof in your textbook actually gives f(x₁) < f(x₂), so f is in fact strictly increasing.
- If f′(x) < 0 for every x in (a, b), f is decreasing on [a, b], and in fact strictly decreasing.
- If f′(x) = 0 for every x in (a, b), f is constant on [a, b].
- The test works one way only. f(x) = x³ is strictly increasing on R, yet f′(0) = 0. A zero derivative at single points does not stop a function from being strictly increasing.
Increasing vs strictly increasing
Board questions usually say 'strictly increasing'. Write the word strictly in your conclusion.
Example 1: increasing on all of R
Show that f(x) = x³ − 6x² + 15x + 4 is strictly increasing on R.
- f′(x) = 3x² − 12x + 15
- Take out 3: f′(x) = 3(x² − 4x + 5)
- Complete the square: x² − 4x + 5 = (x − 2)² + 1
- So f′(x) = 3[(x − 2)² + 1]
- (x − 2)² ≥ 0 for every real x, so (x − 2)² + 1 ≥ 1 > 0.
- Hence f′(x) > 0 for every x in R.
f′(x) > 0 on R, so f is strictly increasing on R.
Example 2: a trigonometric function
Show that f(x) = tan x − x is strictly increasing on (0, π/2).
- f′(x) = sec² x − 1
- Use the identity sec² x − 1 = tan² x, so f′(x) = tan² x.
- For 0 < x < π/2, tan x > 0, so tan² x > 0.
- Hence f′(x) > 0 for every x in (0, π/2).
f is strictly increasing on (0, π/2).
The price of tomatoes in a Chennai market often rises for some weeks and then falls when the new crop arrives. Its graph is increasing on one interval and decreasing on the next.
1 marks: “On R, the function f(x) = e⁻ˣ is: strictly increasing, strictly decreasing, constant, or neither?” f′(x) = −e⁻ˣ < 0 for every x, so f is strictly decreasing on R.
2 marks: “Show that f(x) = 5x + 3 cos x is strictly increasing on R.” f′(x) = 5 − 3 sin x. Since sin x ≤ 1, f′(x) ≥ 5 − 3 = 2 > 0 for every x. So f is strictly increasing on R.
2 marks: “Prove that f(x) = x − log(1 + x) is strictly increasing on (0, ∞).” f′(x) = 1 − 1/(1 + x) = x/(1 + x). For x > 0, both x and 1 + x are positive, so f′(x) > 0. Hence f is strictly increasing on (0, ∞).
f′ positive, graph goes up; f′ negative, graph goes down.