Intro+to+the+derivative+notes.pdf

Assume that x takes values in an interval of the form

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). Assume that x takes values in an interval of the form [ a, b ]. Then the change in x , sometimes called the increment of x , is Δ x = b - a The change in f ( x ) over the interval [ a, b ] is Δ f = f ( b ) - f ( a ) The average rate of change of f ( x ) over the interval [ a, b ] is the difference quotient of f over the interval [ a, b ]. It is given by the following formula: Δ f Δ x = f ( b ) - f ( a ) b - a Geometrically, it represents the slope of the line which passes through the points P ( a, f ( a )) and Q ( b, f ( b )). Such a line is called a secant line of the graph of f . Sometimes, we will be interested in computing the average rate of change of f over the interval [ a, a + h ]. Replacing b by a + h we obtain the following formula: Δ f Δ x = f ( a + h ) - f ( a ) h The units of the change Δ f in f are the units of f ( x ), while the units of the average rate of change of f are units of f ( x ) per unit of x .
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3. Introduction to the derivative: Limits and Continuity 21 Example 3.22. The following table shows approximately daily oil produc- tion by Pemex, Mexico’s national oil company, for 2001 - 2009 ( t = 1 repre- sents the start of 2001). a. Compute the average rate of change of P ( t ) over the period 2002 - 2007 . Interpret the result. b. Which of the following is true? From 2001 to 2008, the one-year average rate of change of oil production by Pemex (A) increased in value. (B) decreased in value. (C) never increased in value. (D) never decreased in value. Solution: a. During the 5-year period [2 , 7] the average rate of P , denoted Δ P is given by Δ P = Δ P Δ t = P (7) - P (2) 7 - 2 = 3 . 2 - 3 . 3 5 = - 0 . 1 5 = = - 0 . 02 million barrels per year = - 20000 barrels per year Interpreting the result: During the period [2 , 7] (that is, 2002 - 2007) the daily oil production by Pemex decreased at an average rate of 20000 barrels of oil per year. b. The values of the one-year average rates are 0 . 2 , 0 . 1 , 0 , 0 , - 0 . 1 , - 0 . 1 Thus (C) is the correct answer.
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22 J. S´ anchez-Ortega 3.6 Instantaneous Rate of Change. Derivative Let f be a function and a a number in the domain of f . The instantaneous rate of change of f ( x ) at x = a is defined as f 0 ( a ) = lim h 0 f ( a + h ) - f ( a ) h The quantity f 0 ( a ) is also called the derivative of f ( x ) at x = a . The units of f 0 ( a ) are the same as the units of the average rate of change: units of f per unit of x . It may happen that the average rates of change [ f ( a + h ) - f ( a )] /h do not approach any fixed number at all as h approaches zero, or that they approach one number on the intervals using positive h , and another one on those using negative h . If this happens, lim h 0 f ( a + h ) - f ( a ) h does not exist, and we say that f is not differentiable at x = a , or f 0 ( a ) does not exist . When the limit does exist, we say that f is differentiable at the point x = a , or f 0 ( a ) exists . Example 3.23. The cost (in dollars) of producing x units of a certain commodity is C ( x ) = 5000 + 10 x + 0 . 05 x 2 . a. Find the instantaneous rate of change of C ( x ) over [100 , 100+ h ] . b. What is the value of the instantaneous rate of change mean?
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