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Rate of Change
at a Point

Understand how average rate of change over an interval leads to the instantaneous rate of change at a single point: the foundation of the derivative.

AP Calculus AB

What Is Rate of Change?

Rate of change measures how one quantity changes in relation to another. If a car travels a distance \( d \) over time \( t \), the rate of change of distance with respect to time is the car's speed.

Slope of a Line
For a linear function, the rate of change is constant and equals the slope \( m \).
Curves are Different
For nonlinear functions, the rate of change varies from point to point.

Average Rate of Change: The Secant Line

The average rate of change of \( f(x) \) on the interval \( [a, b] \) is the slope of the secant line connecting the two points \( (a, f(a)) \) and \( (b, f(b)) \).

x y (a, f(a)) (b, f(b)) Secant line f(x)

Why Average Isn't Enough

The average rate of change gives a big-picture summary over an interval, but it misses the details. Imagine driving from city A to city B in 2 hours, covering 120 miles. Your average speed is 60 mph, but at any given moment you could have been going 40 mph or 80 mph.

The key idea: shrink the interval. As \( b \) gets closer and closer to \( a \), the secant line approaches a line that just touches the curve at one point: the tangent line.

Instantaneous Rate of Change: The Limit

The instantaneous rate of change of \( f(x) \) at \( x = a \) is defined as the limit of the average rate of change as the interval shrinks to zero:

This limit, when it exists, gives the slope of the tangent line to \( f \) at the point \( (a, f(a)) \). This value is called the derivative of \( f \) at \( a \).

From Secant to Tangent

x y a Tangent line Secants approach tangent f(x)

Average vs. Instantaneous

Average Rate of Change Instantaneous Rate of Change
Domain Over an interval \([a, b]\) At a single point \(x = a\)
Geometric Meaning Slope of the secant line Slope of the tangent line
Formula \(\dfrac{f(b)-f(a)}{b-a}\) \(\displaystyle\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}\)
Requires a Limit? No Yes
Analogy Average speed on a trip Speedometer reading

The Difference Quotient

The expression inside the limit is called the difference quotient. It has two equivalent forms, each useful in different contexts:

Form 1: Using \( h \)
$$\frac{f(a+h)-f(a)}{h}$$ Here \( h \) is the horizontal distance from \( a \) to the second point. As \( h \to 0 \), we get the derivative.
Form 2: Using \( x \)
$$\frac{f(x)-f(a)}{x-a}$$ Here \( x \) is the second point directly. As \( x \to a \), the secant becomes the tangent.

How to Find the Instantaneous Rate

1
Write the difference quotient. Substitute \( f(a+h) \) and \( f(a) \) into \( \dfrac{f(a+h)-f(a)}{h} \).
2
Expand and simplify. Expand \( f(a+h) \), combine like terms, and factor out \( h \) from the numerator.
3
Cancel \( h \). After factoring, cancel the \( h \) in the numerator with the \( h \) in the denominator.
4
Take the limit. Let \( h \to 0 \) in the simplified expression to find \( f'(a) \).

Example 1: \( f(x) = x^2 \) at \( x = 3 \)

1
Difference quotient: \( \dfrac{(3+h)^2 - 3^2}{h} = \dfrac{9 + 6h + h^2 - 9}{h} \)
2
Simplify: \( \dfrac{6h + h^2}{h} = \dfrac{h(6 + h)}{h} \)
3
Cancel \( h \): \( 6 + h \)
4
Take the limit: \( \displaystyle\lim_{h \to 0}(6 + h) = \) 6

Visualizing \( f(x) = x^2 \) at \( x = 3 \)

x y 0 1 2 3 4 5 5 12 (3, 9) slope = 6
\( f(x) = x^2 \)

Example 2: \( g(x) = 3x^2 - 2x \) at \( x = 1 \)

1
Evaluate: \( g(1) = 3(1)^2 - 2(1) = 1 \) and \( g(1+h) = 3(1+h)^2 - 2(1+h) \)
2
Expand \( g(1+h) \): \( 3(1 + 2h + h^2) - 2 - 2h = 3 + 6h + 3h^2 - 2 - 2h = 1 + 4h + 3h^2 \)
3
Difference quotient: \( \dfrac{(1 + 4h + 3h^2) - 1}{h} = \dfrac{4h + 3h^2}{h} = 4 + 3h \)
4
Limit: \( \displaystyle\lim_{h \to 0}(4 + 3h) = \) 4

Example 3: Using the Alternate Form

Find the instantaneous rate of change of \( f(x) = \sqrt{x} \) at \( x = 4 \) using \( \displaystyle\lim_{x \to a}\frac{f(x) - f(a)}{x - a} \).

1
Set up: \( \dfrac{\sqrt{x} - \sqrt{4}}{x - 4} = \dfrac{\sqrt{x} - 2}{x - 4} \)
2
Rationalize: Multiply by \( \dfrac{\sqrt{x} + 2}{\sqrt{x} + 2} \) to get \( \dfrac{x - 4}{(x-4)(\sqrt{x}+2)} \)
3
Cancel and evaluate: \( \dfrac{1}{\sqrt{x}+2} \xrightarrow{x \to 4} \dfrac{1}{\sqrt{4}+2} = \) \( \frac{1}{4} \)

When No Tangent Exists

The instantaneous rate of change does not exist at a point if the limit fails. Three common cases:

Corner
Example: \( f(x) = |x| \) at \( x = 0 \). Left-side slope is \(-1\), right-side slope is \(+1\).
Cusp
Example: \( f(x) = x^{2/3} \) at \( x = 0 \). The slope approaches \(\pm\infty\) from each side.
Discontinuity
If \( f \) is not continuous at \( a \), the limit cannot exist. Continuity is necessary for differentiability.

Common Mistakes

✕ Forgetting the Limit
Computing \( \frac{f(a+h)-f(a)}{h} \) and plugging in \( h = 0 \) directly gives \( \frac{0}{0} \). You must simplify first, then take the limit.
✕ Sign Errors in Expansion
When expanding \( f(a+h) \), be careful distributing negatives. Example: \( -(a+h)^2 \neq -a^2 + h^2 \).
✕ Confusing Average with Instantaneous
The average rate of change over \( [a, b] \) is not the derivative. You need the limit as the interval shrinks to zero.

Summary: Rate of Change at a Point

Concept Definition Key Fact
Average Rate of Change \(\dfrac{f(b)-f(a)}{b-a}\) Slope of secant line
Difference Quotient \(\dfrac{f(a+h)-f(a)}{h}\) Before the limit is taken
Instantaneous Rate \(\displaystyle\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}\) Slope of tangent line = \(f'(a)\)
Differentiability Limit exists and is finite Implies continuity

Question 1 of 5

What is the average rate of change of \( f(x) = x^3 \) on the interval \( [1, 3] \)?

Question 2 of 5

Using the limit definition, find \( f'(2) \) for \( f(x) = x^2 + 1 \).

Question 3 of 5

The expression \( \displaystyle\lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \) represents the slope of which line?

Question 4 of 5

At which point is \( f(x) = |x - 2| \) not differentiable?

Question 5 of 5

Which expression is equivalent to \( \displaystyle\lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \)?