A. Rate of Change · Derivatives and Differentiation RulesIncreasing, Decreasing and the Sign of the Rate of Change
A2.1About 65 minutes
Learning goals- I can tell from the sign of the instantaneous rate of change whether a function is increasing or decreasing.
- I can find intervals where the rate of change is positive, negative or zero, numerically and graphically.
- I can describe the rate of change at and near a local maximum or minimum.
- I can explain why a zero rate of change does not always mean a turning point.
Words to know
increasing function
On an interval, f is increasing if f(x₁) < f(x₂) whenever x₁ < x₂: the graph rises from left to right.
decreasing function
On an interval, f is decreasing if f(x₁) > f(x₂) whenever x₁ < x₂: the graph falls from left to right.
local maximum
A point on a graph that is higher than all nearby points: the top of a hill.
local minimum
A point on a graph that is lower than all nearby points: the bottom of a valley.
turning point
A local maximum or local minimum, where the graph changes from increasing to decreasing or the reverse.
Welcome to calculus
In Advanced Functions (MHF4U), in the unit Rates of Change, you estimated how fast a function changes at a single point. You did it by shrinking secants until they looked like a tangent, and you used the difference quotient (f(a + h) − f(a)) ÷ h. In the unit Limits and Rates of Change of this course, you made that idea precise with limits.
This unit asks a bigger question. Instead of finding the rate of change at one point, what if we could find it at every point at once? The answer is a new function called the derivative. By the end of the unit you will find derivatives in a few lines, using rules instead of long limit calculations.
We start with the most useful thing a rate of change tells you: whether the function is going up or down.
The sign of the instantaneous rate of change
Picture yourself skiing along a graph from left to right. The instantaneous rate of change at a point is the slope of the tangent there.
- If the tangent slope is positive, you are going uphill. The function is increasing.
- If the tangent slope is negative, you are going downhill. The function is decreasing.
- If the tangent slope is zero, the tangent is horizontal. You are on flat ground for an instant.
More precisely: a function is increasing on an interval if, for any two x-values in the interval with x₁ < x₂, we have f(x₁) < f(x₂). For a smooth function, a tangent slope that is positive at every point of an interval guarantees that the function is increasing there. A negative slope everywhere on an interval guarantees decreasing.
Turning points: local maximums and minimums
A local maximum is a point higher than all the points near it: the top of a hill. A local minimum is lower than all the points near it: the bottom of a valley.
At a local maximum or minimum of a smooth curve, the tangent is horizontal, so the instantaneous rate of change is 0. Look at the rate on each side:
- Local maximum: the rate changes from positive to negative (uphill, flat, downhill).
- Local minimum: the rate changes from negative to positive (downhill, flat, uphill).
Be careful: a rate of 0 does not always mean a turning point. For y = x³ the tangent at the origin is horizontal, but the graph keeps rising on both sides. What matters is whether the sign of the rate changes.
Zero is a clue, not a verdict
A zero rate of change marks a place to check, not a guaranteed maximum or minimum. Always look at the sign of the rate just to the left and just to the right.
Finding the sign numerically
You can estimate tangent slopes with a centred difference quotient from MHF4U:
slope at x = a ≈ [f(a + 0.01) − f(a − 0.01)] ÷ 0.02
Do this at several x-values and record the signs. The table below does it for f(x) = x³ − 3x.
| x | −2 | −1.5 | −1 | −0.5 | 0 | 0.5 | 1 | 1.5 | 2 |
|---|
| estimated slope | 9.0001 | 3.7501 | 0.0001 | −2.2499 | −2.9999 | −2.2499 | 0.0001 | 3.7501 | 9.0001 |
| sign | + | + | 0 (about) | − | − | − | 0 (about) | + | + |
Worked exampleExample 1: Reading intervals from slope estimates
Use the table for f(x) = x³ − 3x. State the intervals where f is increasing and decreasing, and classify the points at x = −1 and x = 1.
- The slope is positive for x-values to the left of −1, so f is increasing for x < −1.
- The slope is negative between −1 and 1, so f is decreasing for −1 < x < 1.
- The slope is positive again to the right of 1, so f is increasing for x > 1.
- At x = −1 the slope changes from + to −: a local maximum. f(−1) = −1 + 3 = 2.
- At x = 1 the slope changes from − to +: a local minimum. f(1) = 1 − 3 = −2.
Increasing for x < −1 and x > 1; decreasing for −1 < x < 1. Local maximum (−1, 2), local minimum (1, −2).
Worked exampleExample 2: Using the difference quotient to find the sign exactly
For g(x) = −x² + 4x + 5, find the tangent slope at any point x = a. Then find where g is increasing and decreasing, and find its maximum.
- g(a + h) = −(a + h)² + 4(a + h) + 5 = −a² − 2ah − h² + 4a + 4h + 5.
- g(a + h) − g(a) = −2ah − h² + 4h.
- Divide by h: −2a − h + 4. As h → 0 this approaches −2a + 4.
- The slope −2a + 4 is positive when a < 2, zero when a = 2 and negative when a > 2.
- So the slope changes from + to − at x = 2, a maximum. g(2) = −4 + 8 + 5 = 9.
Tangent slope −2a + 4. g is increasing for x < 2, decreasing for x > 2, with a maximum at (2, 9). (This matches the vertex of the parabola, found by completing the square in Grade 11.)
Worked exampleExample 3: A winter temperature
The temperature in Thunder Bay on a January day is modelled by T(t) = −0.05t² + 1.2t − 18 (°C), where t is hours after midnight, 0 ≤ t ≤ 24. When is the temperature rising, when is it falling, and what is the warmest temperature?
- Use the method from Example 2 with a = t: the tangent slope is −0.1t + 1.2. (Check: T(t + h) − T(t) = −0.1th − 0.05h² + 1.2h; divide by h and let h → 0.)
- −0.1t + 1.2 = 0 when t = 12.
- For t < 12 the slope is positive (rising). For t > 12 it is negative (falling).
- T(12) = −0.05(144) + 14.4 − 18 = −7.2 + 14.4 − 18 = −10.8.
Rising from midnight until noon, falling after noon. The warmest temperature is −10.8 °C at noon. At noon the rate of change is 0 °C/h.
Worked exampleExample 4: A function whose slope is always positive
A smooth function has a tangent slope that is positive at every x. Explain how you know the function is increasing, and give an example.
- If every tangent slopes upward, then wherever you are on the graph, moving right takes you up.
- So the graph can never come back down: f(x₁) < f(x₂) whenever x₁ < x₂.
- Example: f(x) = x³ + x. Its tangent slope (by the difference quotient) is 3a² + 1, and 3a² + 1 ≥ 1 > 0 for every a.
The function is increasing everywhere; for example f(x) = x³ + x has slope 3x² + 1, which is always at least 1.
Just for funA tangent line at the top of a hill is perfectly level. That is more than anyone can say about the toboggan hill at your local park in January.
Hands-on activityRuler tangents on a hand-drawn hill
You need: a sheet of paper, a pencil, a ruler (a straight-edged card works too) and a calculator.
- Draw x- and y-axes. Sketch a smooth curve with two hills and one valley between them (like the letter M with rounded tops).
- Lay the ruler so it just touches the curve at one point without cutting across it. That is a tangent. Slide it slowly along the curve from left to right.
- Under the x-axis, write + where the ruler tilts up to the right, − where it tilts down, and 0 where it is level.
- Circle each 0 and label it max or min by looking at the signs on either side.
- Now do it numerically: for f(x) = x³ − 3x, calculate [f(a + 0.01) − f(a − 0.01)] ÷ 0.02 for a = −1.5, 0 and 1.5 and compare with the table in this lesson.
What to look for: the signs should read + 0 − 0 + 0 − from left to right (three zeros: max, min, max). The numerical slopes should be about 3.75, −3.00 and 3.75.
Your turn
1At a certain point the tangent to y = f(x) has slope −4. Is f increasing or decreasing there?
The slope is negative, so f is decreasing there.
2The rate of change of a function is positive for x < 3 and negative for x > 3. What happens at x = 3?
The rate changes from + to −, so there is a local maximum at x = 3.
3The rate of change of f is negative, then 0 at x = −2, then positive. Describe the point at x = −2.
A local minimum (downhill, flat, uphill).
4For f(x) = x² − 6x, the tangent slope at x = a is 2a − 6. Find the intervals of increase and decrease.
2a − 6 < 0 when a < 3 and > 0 when a > 3. Decreasing for x < 3, increasing for x > 3, with a minimum at (3, −9).
5Estimate the tangent slope of f(x) = x³ at x = 2 using [f(2.01) − f(1.99)] ÷ 0.02.
f(2.01) = 8.120601 and f(1.99) = 7.880599. Difference 0.240002 ÷ 0.02 = 12.0001, so about 12.
6The rate of change of y = x³ is 0 at x = 0. Is (0, 0) a turning point? Explain.
No. The slope is positive on both sides of 0 (3a² > 0 for a ≠ 0), so the graph keeps rising through the origin.
7A hot-air balloon's height has a rate of change of 0 m/s at t = 40 s, positive before and negative after. What is happening at t = 40 s?
The balloon reaches its greatest height (a local maximum) and starts to come down.
More practice: warm-up
1If a function is decreasing on an interval, what sign do its tangent slopes have there?
2What is the slope of the tangent at a local minimum of a smooth curve?
3The slope goes from − to + at x = 5. Maximum or minimum?
4A linear function has slope 2. Where is it increasing?
Everywhere (its rate of change is always 2).
5A constant function f(x) = 7 has what instantaneous rate of change?
0 everywhere: the graph is a horizontal line.
More practice: on your own
1For f(x) = 3x² + 12x − 1, find the tangent slope at x = a with the difference quotient, then find the intervals of increase and decrease.
f(a + h) − f(a) = 6ah + 3h² + 12h; divide by h: 6a + 3h + 12 → 6a + 12. Zero at a = −2. Decreasing for x < −2, increasing for x > −2; minimum f(−2) = 12 − 24 − 1 = −13.
2A snowboarder's height above the bottom of a hill (in metres) is h(t) = −2t² + 8t + 40 during a jump, t in seconds. When is the height increasing?
Slope −4t + 8 (difference quotient). Positive when t < 2. The height increases for 0 ≤ t < 2 s, peaks at h(2) = 48 m, then decreases.
3Estimate the sign of the rate of change of f(x) = x⁴ − 2x² at x = −0.5 and x = 0.5 using h = 0.01. What happens at x = 0?
f′ is about 4a³ − 4a: at −0.5 about +1.5 and at 0.5 about −1.5 (centred estimates give 1.4998 and −1.4998). The slope changes from + to −, so (0, 0) is a local maximum.
4Sketch (in words) a smooth function that is increasing for x < 1, decreasing for 1 < x < 4 and increasing for x > 4.
A hill with its top at x = 1 (local max) and a valley with its bottom at x = 4 (local min), like a cubic with positive leading coefficient. Any such curve is correct if it turns at x = 1 and x = 4.
5True or false: if f′ is zero at x = a then f has a local maximum or minimum at a. Give a reason.
False. y = x³ has a horizontal tangent at 0 but no turning point, because the slope does not change sign.
Challenge
1For f(x) = x³ − 12x, the tangent slope is 3a² − 12. Find all local maximums and minimums and the intervals of increase and decrease.
3a² − 12 = 0 gives a = ±2. For a < −2, slope > 0; for −2 < a < 2, slope < 0; for a > 2, slope > 0. Local max (−2, 16), local min (2, −16). Increasing for x < −2 and x > 2; decreasing for −2 < x < 2.
2Can a smooth function have a local maximum at x = 1 and a local maximum at x = 3 with no local minimum between them? Explain.
No. After the maximum at x = 1 the function decreases. To reach another maximum at x = 3 it must start increasing again, so the slope changes from − to + somewhere between 1 and 3: a local minimum.
3The function f(x) = x|x| has tangent slope 2|x|. Is f increasing on all real numbers? Is there a turning point?
The slope 2|x| is positive except at x = 0, where it is 0. The function is increasing everywhere (for x₁ < x₂, x₁|x₁| < x₂|x₂|) and has no turning point; like x³, the flat point at 0 is not a max or min.
Quiz
1. On an interval where every tangent slope of a smooth function is negative, the function is:
2. The rate of change of f is + for x < 2 and − for x > 2. What does f have at x = 2?
3. The tangent slope of g(x) = x² + 10x at x = a is 2a + 10. Where is g decreasing?
4. Which function has a horizontal tangent at x = 0 but no local maximum or minimum there?
5. The centred estimate [f(1.01) − f(0.99)] ÷ 0.02 for a function is −5.02. What does this suggest at x = 1?
6. A function's slope changes from negative to positive at x = −3. Which describes the point?
Big idea (A2.1): the sign of the instantaneous rate of change (the tangent slope) tells you where a function goes up and where it goes down. Where the sign changes, there is a local maximum or minimum.
How to help: Draw any wavy curve and run a pencil along it as a tangent. Ask your teen "+, − or 0 here?" at several points, then ask where the hills and valleys are. Ask them to explain why y = x³ is flat at 0 but has no hill there.
Common difficulties:
- Mixing up the value of the function with its rate of change (a function can be negative and increasing at the same time).
- Assuming every zero slope is a maximum or minimum. Teens should always check the signs on both sides.
- Writing intervals with the y-values instead of the x-values. Intervals of increase and decrease are always stated in terms of x.