Rate of Change and Derivatives

Rate of Change and Derivatives — Moosa Academy

A car that covers 200 km in two hours averaged 100 km/h, but it was almost never travelling at exactly that speed. The average rate of change answers the first question; letting the two measuring points slide together answers the second — and produces the derivative.

Concept Average rate of change

\frac{f(x_2) - f(x_1)}{x_2 - x_1}

The average rate is the slope of the secant — the straight line joining two points on the curve.

The instantaneous rate is the slope of the tangent at a single point.

Bring $x_2$ closer and closer to $x_1$ . The secant pivots until, in the limit, it becomes the tangent — and that limiting slope is the derivative.

Concept What the sign tells you

Rate of change	Behaviour	Meaning
Positive	Rising	The function is increasing
Negative	Falling	The function is decreasing
Zero	Level	A peak, a valley or an inflection

This is why setting $f'(x) = 0$ is the standard route to finding peaks and valleys: at those points the function has stopped moving in one direction and not yet started in the other.

Example An average rate

Find the average rate of change of $f(x) = x^2$ between $x = 1$ and $x = 3$ .

f(1) = 1

and

f(3) = 9

\dfrac{9 - 1}{3 - 1} = \dfrac{8}{2}

⟹ average rate of change = 4

The curve was not changing at a rate of 4 throughout — that is simply the overall figure across the interval.

Example An instantaneous rate

Find the instantaneous rate of change of $f(x) = x^2$ at $x = 3$ .

Differentiate:

f'(x) = 2x

Substitute:

f'(3) = 6

⟹ instantaneous rate of change = 6

Compare this with the average of 4 over $[1,\ 3]$ . The curve steepens as $x$ grows, so its rate at the right-hand end exceeds the average across the whole stretch.

Example Locating peaks and valleys

Find the turning points of $f(x) = -x^3 + 3x$ .

f'(x) = -3x^2 + 3

Set to zero:

-3x^2 + 3 = 0

, so

x^2 = 1

and

x = \pm 1

f(1) = -1 + 3 = 2

f(-1) = 1 - 3 = -2

To the left of

x = 1

the derivative is positive and to the right negative, so this is a peak.

⟹ peak at (1, 2), valley at (−1, −2)

Summary

The average rate of change is the slope of the secant between two points.
Sliding the two points together turns the secant into the tangent, whose slope is the derivative.
A positive derivative means rising, negative means falling, zero means a turning point.
Solving f′(x) = 0 is the standard method for locating peaks and valleys.