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# Mark Rscribe

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### Mark Rscribe

1. 1. We started today's class by deriving the difference quotient. It is the average rate of change of y with respect to x the given interval Graphically, it is the slope of the secant line connecting points P and Q. Title: Sep 27-7:52 PM (1 of 10)
2. 2. The slope of the secant line through points P and Q is given by this formula: Title: Sep 27-8:14 PM (2 of 10)
3. 3. We then moved on to do Example 1 on the book which is on page 90. Let the function f be defined by a) Find the average rate of change of f over the interval Title: Sep 27-8:24 PM (3 of 10)
4. 4. b) Find the equation of the corresponding secant line. Title: Sep 27-8:41 PM (4 of 10)
5. 5. c) Plot the graph of f and the secant line Title: Sep 27-8:48 PM (5 of 10)
6. 6. We then talked about the instantaneous rate of change. If the average rate of change has a limiting value as the interval decreases in size then it is called the instantaneous rate of change of outputs with respect to inputs. Graphically, it is represented by a tangent line. Title: Sep 27-8:57 PM (6 of 10)
7. 7. This is the traditional notation for the limiting value which is read as `` the limit as x approaches zero of the difference quotient``. Title: Sep 27-9:05 PM (7 of 10)
8. 8. We then worked on another example. 2 Let f(x)=x a) Find the slope of the tangent line to the graph of f at (2,12) Now, as h gets smaller 3h approaches zero and the limiting value is 12. Title: Sep 27-9:11 PM (8 of 10)
9. 9. b) Find the equation of the tangent line The tangent line passes through the point (2,12) with the slope of 12. Title: Sep 27-9:28 PM (9 of 10)
10. 10. Thanks everyone that is the scribe for today. Remember to study tonight for tomorrow`s test on the first unit. Our homework for tonight is Exercise 2.2 questions number 2,10,12,24 and all odd numbers. Title: Sep 27-9:33 PM (10 of 10)