1. AP CALCULUS POWERPOINT- FRQ MONICA POPA

2. 2005 FRQ #2- CALCULATOR

3. PART (A)

4. PART (B)

5. PART (C)

6. PART (D) METHOD 1

7. PART (D) CALCULATOR Enter equation in SOLVER method.Find the value of x in the interval [0,6] which is the critical number.

8. PART (D) METHOD 2 Since the function to find critical points is S(t)—R(t)=0, then S(t)=R(t). This means the two functions are equal to each other and the value of t, the critical point, can be found graphically as well as algebraically. The critical point is the intersection point of the two graphs. Solution: Graph the two functions in the graphing calculator and set an appropriate window  [0,6]. The two functions intersect and to find the intersection point press 2 nd, TRACE, 5, ENTER, ENTER and the calculator will find the x and y values of the intersection point. Answer: Intersection point/critical point  t=5.118.

9. PART (D) CONTINUED t Y(t) 0 2500 5.118 2492.369 6 2493.277  Answers second part of question: what is the minimum value?

10. PART (D) CONTINUED

11. PART (D) CALCULATOR Value of Y(t) when t=5.118. Value of Y(t) when t=6.

12. The College Board- http://apcentral.collegeboard.com/apc/mem bers/exam/exam_information/1997.html http://apcentral.collegeboard.com/apc/mem bers/exam/exam_information/1997.html http://apcentral.collegeboard.com/apc/public/r epository/_ap05_sg_calculus_ab_46569.pdf CITATIONS