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2.1 Defining Average and Instantaneous Rates of Change at a Point

Syllabus
2020
Topic
2.1
Level
—

Learning objectives

—CHA-2.A—Determine average rates of change using difference quotients• CHA-2.A Determine average rates of change using difference quotients.• CHA-2.A.1 The difference quotients (f(a + h) − f(a))/h and (f(x) − f(a))/(x − a) express the average rate of change of a function over an interval.• Enduring understanding CHA-2: Derivatives allow us to determine rates of change at an instant by applying limits to knowledge about rates of change over intervals.—CHA-2.B—Represent the derivative of a function as the limit of a difference quotient• CHA-2.B Represent the derivative of a function as the limit of a difference quotient.• CHA-2.B.1 The instantaneous rate of change of a function at x = a can be expressed by lim h→0 (f(a + h) − f(a))/h or lim x→a (f(x) − f(a))/(x − a), provided the limit exists. These are equivalent forms of the definition of the derivative and are denoted f′(a).• Enduring understanding CHA-2: Derivatives allow us to determine rates of change at an instant by applying limits to knowledge about rates of change over intervals.

Average Rate of Change as a Difference Quotient

Average rate of change measures output change per unit of input change across an interval. Geometrically, it is the slope of the secant line through the two endpoint points on the graph.

\text{Average rate on }[a,b]=\frac{f(b)-f(a)}{b-a}=\frac{f(a+h)-f(a)}{h},\qquad h=b-a\ne0

For f(x)=x2f(x)=x^2 on [2,5][2,5], the output change is f(5)−f(2)=25−4=21f(5)-f(2)=25-4=21 and the input change is 5−2=35-2=3. The average rate of change is 21/3=721/3=7, so the secant line through (2,4)(2,4) and (5,25)(5,25) has slope 77.

This is an average over the whole interval, not necessarily the rate at either endpoint. Its units are output units per input unit, and the order of subtraction must match in numerator and denominator.

Instantaneous Rate from the Derivative Limit

The instantaneous rate of change at x=ax=a is the limit of average rates over intervals whose second endpoint approaches aa. If this limit exists, it is the derivative f′(a)f'(a) and the slope of the tangent line at (a,f(a))(a,f(a)).

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

For f(x)=x2f(x)=x^2 at a=3a=3, f(3+h)−f(3)h=(3+h)2−9h=6h+h2h=6+h\dfrac{f(3+h)-f(3)}{h}=\dfrac{(3+h)^2-9}{h}=\dfrac{6h+h^2}{h}=6+h for h≠0h\ne0. Taking h→0h\to0 gives f′(3)=6f'(3)=6. Thus the instantaneous rate and tangent slope at x=3x=3 are 66.

Do not substitute h=0h=0 into the original quotient; it is undefined there. Simplify for nonzero hh, then take the limit. If the two-sided limit does not exist, the function is not differentiable at aa.