Difficulty: Introductory
Prerequisites: Algebra (function notation, slope of a line, evaluating polynomials and trigonometric functions)
This section introduces the central idea that calculus is built on: measuring how a function changes at a single point rather than over an interval. You have already worked with slopes of straight lines. Here, you extend that idea to curves by using secant lines that get closer and closer to the point of interest. The slope of the tangent line, the thing those secant slopes approach, turns out to be the foundation of the derivative, which you will formalise in the next few sections. If you are comfortable evaluating functions and computing the slope between two points, you have everything you need.
The average rate of change of a function over an interval is the slope of the secant line connecting the interval's endpoints. As the interval shrinks toward a single point, that secant slope approaches the slope of the tangent line, which tells you the function's instantaneous rate of change. This limiting idea is the doorstep of the derivative.
Average rate of change
The ratio [f(b) − f(a)] / (b − a), giving the overall change in a function's output per unit change in input over the interval [a, b]. In simple terms, this is the slope of the straight line connecting two points on a curve.
Secant line
A straight line passing through two distinct points on the graph of a function. In simple terms, think of it as a "chord" of the curve that approximates the curve's behaviour between those two points.
Tangent line
The line that just touches the curve at a single point and matches the curve's direction there. In simple terms, it is the line you would see if you zoomed in on the curve at that point until the curve looked straight.
Slope of the tangent line
The value the secant-line slopes approach as the two points on the curve are squeezed together toward the point of tangency. In simple terms, it is the instantaneous rate of change of the function at that point, and it becomes the derivative once limits are formalised.
Instantaneous velocity
The speed and direction of a moving object at a single instant, obtained by taking the limit of average velocities over shorter and shorter time intervals. Think of it as what your speedometer reads at one exact moment.
Computing the average rate of change
Pick two x-values, a and b, on the function's domain.
Evaluate the function at both: f(a) and f(b).
The average rate of change is the difference quotient: [f(b) − f(a)] / (b − a).
Geometrically, this is the slope of the secant line through (a, f(a)) and (b, f(b)).
Reading it from a graph
Identify the two marked points (often highlighted in red or labelled).
Read their coordinates from the axes.
Apply the slope formula: rise / run = (y₂ − y₁) / (x₂ − x₁).
A positive result means the function is increasing on average over that interval; negative means decreasing.
Working with polynomial functions
For f(x) = x² + 5x − 10 on the interval [2, 4]: evaluate f(2) = 4 + 10 − 10 = 4, and f(4) = 16 + 20 − 10 = 26. The secant slope is (26 − 4) / (4 − 2) = 11.
On [2, 3]: f(3) = 9 + 15 − 10 = 14, so slope = (14 − 4) / (3 − 2) = 10.
On [1, 2]: f(1) = 1 + 5 − 10 = −4, so slope = (4 − (−4)) / (2 − 1) = 8.
Working with trigonometric functions
For f(t) = 2 + cos t on [0, π/4]: f(0) = 3 and f(π/4) = 2 + √2/2 ≈ 2.707. The average rate of change is (2.707 − 3) / (π/4 − 0) ≈ −0.373.
On [−π/4, π/4]: f(−π/4) = 2 + √2/2 ≈ 2.707 and f(π/4) ≈ 2.707. Since both endpoints give the same value, the average rate of change is 0. The function is symmetric about t = 0 here, so the rise cancels out.
For f(x) = −2 tan(x) − 3, the same procedure applies. Evaluate at both endpoints and divide by the interval length.
The squeeze strategy
Compute the secant slope on an interval just to the right of the point (e.g. [2, 3]).
Compute the secant slope on an interval just to the left (e.g. [1, 2]).
Average those two slopes to estimate the tangent slope at the shared endpoint.
Example with f(x) = x² + 5x − 10 at x = 2
Secant slope on [2, 3] = 10 (from the calculation above).
Secant slope on [1, 2] = 8.
Estimated tangent slope = (10 + 8) / 2 = 9.
This estimate improves if you use intervals that are narrower and more symmetric around x = 2.
Example with f(x) = −2 tan(x) − 3 at x = π/6
Compute the secant slope on [π/6, π/3] and on [0, π/6].
Average the two to estimate the tangent slope at x = π/6.
Why it works
If the curve is smooth (no sharp corners, no breaks), the secant slopes from the left and from the right both approach the same value as the interval shrinks. The average of a left-side and a right-side secant is a reasonable first estimate when you do not yet have the machinery of limits.
Using an h-parameter
For f(x) = x² + 3x, the secant slope joining P(2, f(2)) and Q(2 + h, f(2 + h)) is computed by expanding and simplifying.
f(2) = 10. f(2 + h) = (2 + h)² + 3(2 + h) = 4 + 4h + h² + 6 + 3h = 10 + 7h + h².
Secant slope = [(10 + 7h + h²) − 10] / h = (7h + h²) / h = 7 + h.
As h → 0, the secant slope approaches 7, so the tangent slope at x = 2 is 7.
Point-slope form
Once you know the tangent slope m at a point (a, f(a)), the tangent line equation is y − f(a) = m(x − a), which rearranges to y = m(x − a) + f(a).
Worked example: f(x) = x² + 3x at x = 2
f(2) = 4 + 6 = 10, so the point is (2, 10).
The tangent slope (from the h-parameter work above) is 7.
Tangent line: y − 10 = 7(x − 2), which simplifies to y = 7x − 4.
\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}\text{Secant slope} = \frac{f(x + h) - f(x)}{h}\text{Tangent slope} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\text{Tangent line equation:} \quad y - f(a) = m(x - a)Average rate of change is exactly what you compute when you work out your average speed on a road trip: total distance divided by total time. The tangent-line idea is how a speedometer works conceptually, giving your speed at one instant rather than over the whole journey. Engineers use the same principle whenever they need to know how quickly a quantity (temperature, pressure, population, stock price) is changing right now, not just on average.
Students often confuse the secant line with the tangent line. The secant passes through two points on the curve; the tangent touches at one point and matches the curve's direction there.
A common mistake is to compute (a − b) in the denominator instead of (b − a). The order matters: the denominator must match the order of the numerator.
Some students assume the average of two secant slopes always gives the exact tangent slope. It does not. It gives an estimate that improves as the intervals narrow, but equality requires taking a limit.
Students sometimes think a tangent line can only touch the curve at one point overall. A tangent line is defined locally, at the point of tangency; it may cross the curve elsewhere.
⚠️ Expect to compute secant slopes for polynomial and trigonometric functions by hand. Know the values of cos(0), cos(π/4), tan(0), tan(π/6), tan(π/3) without a calculator.
⚠️ The "use parts (b) and (c) to estimate the tangent slope" pattern appears repeatedly. The expected method is to average the two neighbouring secant slopes.
⚠️ Be ready to simplify the difference quotient [f(x + h) − f(x)] / h algebraically and then let h → 0. This is the bridge between Section 1.4 and the formal definition of the derivative in Section 2.1.
⚠️ Tangent line equations (point-slope form) are a standard exam question. You need the slope and the point; mistakes usually come from arithmetic errors in evaluating f(a).
True or false: the average rate of change of f(x) over [a, b] equals the slope of the secant line through (a, f(a)) and (b, f(b)). True.
Fill in the blank: as h → 0, the difference quotient [f(x + h) − f(x)] / h gives the slope of the ______ line. tangent
True or false: if f(−π/4) = f(π/4), the average rate of change of f over [−π/4, π/4] is zero. True.
Fill in the blank: for f(x) = x² + 3x, the secant slope from (2, f(2)) to (2 + h, f(2 + h)) simplifies to ______ + h. 7
True or false: you need a limit to find the exact tangent slope; averaging two secant slopes only gives an estimate. True.
Q: Find the average rate of change of f(x) = x² + 5x − 10 on the interval [2, 4].
A: f(2) = 4, f(4) = 26. Average rate of change = (26 − 4) / (4 − 2) = 11.
Q: For f(x) = x² + 3x, find the slope of the secant line joining P(0, f(0)) and Q(4, f(4)).
A: f(0) = 0, f(4) = 16 + 12 = 28. Slope = (28 − 0) / (4 − 0) = 7.
Q: For f(x) = x² + 3x, simplify the secant slope from P(2, f(2)) to Q(2 + h, f(2 + h)) and then estimate the tangent slope at x = 2.
A: Secant slope = 7 + h. As h → 0, the tangent slope is 7.
Q: Write the equation of the tangent line to f(x) = x² + 3x at x = 2.
A: The point is (2, 10), the slope is 7, so y = 7(x − 2) + 10 = 7x − 4.
Q: What is the average rate of change of f(t) = 2 + cos t over [−π/4, π/4], and why?
A: Zero. cos t is symmetric about t = 0, so f(−π/4) = f(π/4), making the numerator zero.
This connects directly to the formal definition of the derivative (Section 2.1), which replaces the estimation approach here with a proper limit. The velocity interpretation resurfaces in related rates (Section 3.9) and in physics-flavoured optimisation problems. The secant-to-tangent idea also underpins numerical methods like Newton's method (Section 4.8), where tangent lines are used to approximate roots.
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