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Clicker Question 1 Given: f(x) = 1 + x + x2/2 + x3/3! + x4/4!, g(x) = x - x3/3! + x5/5!, and h(x) = (x –1) – (x –1)2/2 + (x –1)3/3 – (x –1)4/4.

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Presentation on theme: "Clicker Question 1 Given: f(x) = 1 + x + x2/2 + x3/3! + x4/4!, g(x) = x - x3/3! + x5/5!, and h(x) = (x –1) – (x –1)2/2 + (x –1)3/3 – (x –1)4/4."— Presentation transcript:

1 Clicker Question 1 Given: f(x) = 1 + x + x2/2 + x3/3! + x4/4!, g(x) = x - x3/3! + x5/5!, and h(x) = (x –1) – (x –1)2/2 + (x –1)3/3 – (x –1)4/ (x –1)5/5 – (x –1)6/6. A calculator would compute sin( ) by computing: A. f(0.435) D. g(0.435) + 12 B. g(0.435) E. g( ) C. h(0.435)

2 Clicker Question 2 Given: f(x) = 1 + x + x2/2 + x3/3! + x4/4!, g(x) = x - x3/3! + x5/5!, and h(x) = (x –1) – (x –1)2/2 + (x –1)3/3 – (x –1)4/ (x –1)5/5 – (x –1)6/6. A calculator would compute e6.435 by computing: A. f(0.435) D. 6(2.7183) + f(0.435) B. g(0.435) E. (2.7183)6 f(0.435) C. h(0.435)

3 Clicker Question 3 Given: f(x) = 1 + x + x2/2 + x3/3! + x4/4!, g(x) = x - x3/3! + x5/5!, and h(x) = (x –1) – (x –1)2/2 + (x –1)3/3 – (x –1)4/ (x –1)5/5 – (x –1)6/6, and ln(10)  A calculator would compute ln(6.435) by computing: A. f(6.435) D h(0.6435) B. g(6.435) E h(0.6435) C. h(6.435)

4 End-of-Semester Schedule
Hand-in #4 is due tomorrow (Thurs 12/12) at 4:45. It will be returned outside my door, together with your clicker score and average coming into the final, sometime on Monday (I’ll you when ready). Office hours: Thursday 3:15-4:45 (as usual), Monday 11-noon and 3-5, Tues 10-noon, Wed 2-4. Final exam is 6-9 pm Wed Dec 18. Standard format. You may bring 2 reference sheets (4 sides). It will cover the whole course with slight emphasis on the infinite series (as that has not yet been tested).


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