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Section 2.2 Average and Instantaneous Rate of Change The Derivative of a Function at a Point 3.1.

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Presentation on theme: "Section 2.2 Average and Instantaneous Rate of Change The Derivative of a Function at a Point 3.1."— Presentation transcript:

1 Section 2.2 Average and Instantaneous Rate of Change The Derivative of a Function at a Point 3.1

2 You are based in Indonesia, and you monitor the value of the US Dollar on the foreign exchange market very closely during a rather active five-day period. Suppose you find that the value of one US dollar can be well approximated by the function (The rupiah is the Indonesian currency), where t is time in days. (t = 0 represents the value of the Dollar at noon on Monday) a) What was the value of the Dollar at noon on Tuesday?

3 b)According to the graph, when was the value of the Dollar rising most rapidly? Monday at Noon

4 c) Compute the average rate of change of R(t) over the interval [1, 1+h] for h = 1, h = 0.01, h = 0.001, and h = 0.0001 HUH?

5 [1, 1 + 1] = [1, 2] [1, 1 + 0.01] = [1, 1.01] [1, 1 + 0.001] = [1, 1.001] [1, 1 + 0.0001] = [1, 1.0001] So what are the conclusions from this?

6 AS h -> 0……. h = 1, h = 0.01, h = 0.001, and h = 0.0001 The slope of the secant line approaches 300…. 200, 299, 299.9, 299.99 Based upon this conclusion, we can say: Back to the problem…….. On Tuesday, the Rupiah was increasing in value at A rate of 300 rupiahs per day. General Conclusion…..

7 A look at the definition of the first derivative at a point in action Note: Scroll down for the second

8 Let f(x) = ln x a) Find the average rate of change of f between x = 0.99 and x = 1. b) Find the average rate of change of f between x = 1 and x = 1.01. c) Explain why the answer in (a) is to large of an estimate for f (x) and the answer in (b) is too small of an estimate for f (x).

9 Let f(x) = ln x a)Find the average rate of change of f between x = 0.99 and x = 1 b)Find the average rate of change of f between x = 1 and x = 1.01

10 Let f(x) = ln x a) The average rate of change between x = 0.99 and x = 1 is 1.005 b) The average rate of change between x = 1 and x = 1.01 is 0.995 c) Explain why the answer in (a) is to large of an estimate for f (x) and the answer in (b) is too small of an estimate for f (x).

11 Given the graph below, where does the derivative NOT exist? X X X X X X X X X


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