# 3.3 - MATH 1081  Wednesday, February 2   ...

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Unformatted text preview: MATH 1081  Wednesday, February 2    Chapter 3 – Section 3    RATES OF CHANGE    Homework #3 (due 2/7):   Section 3.2 #6, 20, 26                  Section 3.3 #8, 30, 34    Clicker Check­in:  Choose any letter to check in now.  Before we begin Section 3.3, let’s look more closely at the last  example from last lecture.    ⎧ x 2 + 1 if x < 0 ⎪ Is the function  g ( x ) = ⎨ −1 if x = 0  continuous at  x = 0 ?  ⎪ ex if x > 0 ⎩   If not, what is happening at the point  x = 0 ?  • Is the point just in the “wrong” place (removable  discontinuity)?  • Do the 2 one‐sided limits not agree (jump discontinuity)?  • Do either of the one‐sided limits go to infinity (asymptote)?       Suppose you are travelling along a highway and at 1pm you are  at mile marker 37.  Later at 2:30pm, you see that you are at  mile marker 142 (on the same highway in the same state).    What was your average rate of velocity for that leg of the trip?        Now, suppose that you are in a car at a stoplight.  When the  light turns green, you begin to drive along a road so that your  distance (in feet) from the light  t  seconds later is given by  s ( t ) = 3t 2 .    What is your average rate of velocity over the first 15 seconds  after you leave the stoplight?  Definition:  Average Rate of Change  The average rate of change of  f ( x )  with respect to  x  from  x = a  to  x = b  is  f (b ) − f ( a )         .  b−a     Sometimes a ratio such as this is called a difference quotient.  Note that this is the same as the slope of the line through the  points ( a, f ( a ))  and ( b, f ( b )) .    f (b ) − f ( a )     b−a     Example:  Determine the average rate of change of the    function over the given interval.    y = 7 x − 9   between  x = −3 and  x = 5   1.         2. f ( x ) = −4 x 2 − 6  between  x = 2  and  x = 6       Back to the car at the stoplight…  Suppose that you are in a car at a stoplight.  When the light  turns green, you begin to drive along a road so that your  distance (in feet) from the light  t  seconds later is given by  s ( t ) = 3t 2 .      How can we determine the speed of the car at one precise  moment (like 10 seconds after leaving the stoplight)?    Certainly we can estimate it with an average velocity over  some time interval containing  t = 10 , but which estimate is the  best one?    Here is where the idea of a limit comes in.  For the best  estimate, we can find the average velocity of smaller and  smaller intervals of time;   like from  t = 10  to  t = 10.1 and then  t = 10  to  t = 10.05  and so on.    We can’t compute the rate of change if we only have one point,  so the length of the interval never equals 0, but we want it to  become smaller and smaller.    So, we want to compute  f (b ) − f ( a )               lim .  b→ a b−a   Looking at the graph, we are looking for the limiting value of  the slope of the line as the two points ( a, f (a ))  and ( b, f (b ))  get  closer and closer together.        For the purpose of computation, we will make a slight  adjustment to the expression a write the following:    Definition: Instantaneous Rate of Change  The instantaneous rate of change of a function  f ( x )  at  x = a   is given by  f (a + h) − f (a)           lim   h→ 0 h provided that the limit exists.      If we visualize the graph of  f ( x ) , we are computing the slope  between the points ( a, f ( a ))  and ( a + h, f ( a + h ))  and then  making two points get closer and closer by  h → 0 .          Example:  Determine the instantaneous rate of change of the    function at the given point.    y = 7 x − 9   at   x = −3  1.       2. f ( x ) = −4 x 2 − 6   at   x = 2         3. g ( x ) = 3x − 2   at   x = 1    Clicker Check­out: Choose any letter to check out now.      Tomorrow in recitation:  Workshop #2    Next time:  Section 3.4    ...
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