AP CALCULUS POWERPOINT- FRQ MONICA POPA
2005 FRQ #2- CALCULATOR
PART (A)
PART (B)
PART (C)
PART (D) METHOD 1
PART (D) CALCULATOR Enter equation in SOLVER method.Find the value of x in the interval [0,6] which is the critical number.
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.
PART (D) CONTINUED t Y(t) Answers second part of question: what is the minimum value?
PART (D) CONTINUED
PART (D) CALCULATOR Value of Y(t) when t= Value of Y(t) when t=6.
The College Board- bers/exam/exam_information/1997.html bers/exam/exam_information/1997.html epository/_ap05_sg_calculus_ab_46569.pdf CITATIONS