Individual Project ~ FRQ Presentation 2002 AP Calculus AB #1 Marija Jevtic.

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Extreme Values of Functions

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Presentation transcript:

Individual Project ~ FRQ Presentation 2002 AP Calculus AB #1 Marija Jevtic

There are no points of intersection, but we are given an interval from to x = 1 There are no points of intersection, but we are given an interval from to x = 1

In order to find the area of a region enclosed by two graphs, we must use the following formula where f(x) represents the upper function, and g(x) represents the lower function, a is the upper limit ( ) b is the lower limit (1) The answer, when the equation is plugged in the calculator is 1.22

y = 4 In order to find the volume of the solid revolved around y = 4 we must use the Washer Method R = r = When all the information is plugged into the formula, we get: The R and r are determined in accordance to the line around which the graph revolves (y = 4). R is the outer ring (further from the line), while r is the inner ring, meaning that it is closer to the given line. Also, the specific line must be included in the formula if it is anything but y = 0. Answer:

First find the derivative of h(x) and set it equal to zero in order to find the critical values An absolute minimum and maximum occur at either the critical point or at one of the endpoints, meaning we must find h(x) values for all of them and test which is the lowest one x = Absolute minimum value: h( ) = Absolute maximum value: h(1) = 2.718

References Microsoft Word (for equations)

THANK YOU! Hope you enjoyed!