Net Change, Average Rate of Change, and Difference Quotient Flashcards

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Vocabulary flashcards covering core definitions, formulas, step-by-step procedures, and specific numerical examples for Net Change, Average Rate of Change, Difference Quotient, and Graph Reading concepts from the lecture notes.

Last updated 4:30 PM on 8/31/26
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13 Terms

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Net Change

The difference between what something ended at and what it started at, defined as f(b)f(a)f(b) - f(a) given a starting point (a,f(a))(a, f(a)) and an ending point (b,f(b))(b, f(b)).

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Average Rate of Change

The ratio of the net change of a function to the difference between its endpoints, calculated using the formula f(b)f(a)ba\frac{f(b) - f(a)}{b - a}.

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Difference Quotient

An expression for the average rate of change formulated by writing the second point in terms of the first point (a+h,f(a+h))(a+h, f(a+h)), defined as f(a+h)f(a)h\frac{f(a+h) - f(a)}{h}.

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Graph Height f(a)f(a)

The height above or below the x-axis at the input value aa on the graph of a function.

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Graph Endpoint Conventions

Graphical notation where visible dots mean the graph ends at those points in the picture, and no dots mean the graph continues forever.

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Domain (Page 6 Graph)

The set of x-values represented on the example graph, given as the interval [1,5][1, 5].

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Range (Page 6 Graph)

The set of y-values represented on the example graph, given as the interval [2,3][-2, 3].

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Net Change Calculation Example (f(t)=3t2+2t+1f(t) = 3t^2 + 2t + 1)

The evaluation of net change between t=1t = 1 and t=4t = 4, given by f(4)f(1)=576=51f(4) - f(1) = 57 - 6 = 51.

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Average Rate of Change Calculation Example (f(t)=3t2+2t+1f(t) = 3t^2 + 2t + 1)

The evaluation of average rate of change from t=1t = 1 to t=4t = 4, given by f(4)f(1)41=513=17\frac{f(4) - f(1)}{4 - 1} = \frac{51}{3} = 17.

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Average Rate of Change Example (f(x)=x32xf(x) = x^3 - 2x)

The evaluation of average rate of change from x=2x = -2 to x=2x = 2, given by f(2)f(2)2(2)=4(4)4=84=2\frac{f(2) - f(-2)}{2 - (-2)} = \frac{4 - (-4)}{4} = \frac{8}{4} = 2.

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Step 1 of Difference Quotient for f(x)=32x2f(x) = 3 - 2x^2

Evaluating and expanding f(a+h)=32(a+h)2=32(a2+2ah+h2)=32a24ah2h2f(a+h) = 3 - 2(a+h)^2 = 3 - 2(a^2 + 2ah + h^2) = 3 - 2a^2 - 4ah - 2h^2.

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Step 3 of Difference Quotient for f(x)=32x2f(x) = 3 - 2x^2

Subtracting f(a)f(a) from f(a+h)f(a+h), which yields f(a+h)f(a)=(32a24ah2h2)(32a2)=4ah2h2f(a+h) - f(a) = (3 - 2a^2 - 4ah - 2h^2) - (3 - 2a^2) = -4ah - 2h^2.

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Step 4 of Difference Quotient for f(x)=32x2f(x) = 3 - 2x^2

Dividing the simplified expression f(a+h)f(a)f(a+h) - f(a) by hh, which yields 2h24ahh=2h4a\frac{-2h^2 - 4ah}{h} = -2h - 4a.