Examples of the Accumulation Function (ANSWERS) dy dx. This new function now passes through (0,2). Make a sketch of your new shifted graph.

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1 Eamples of the Accumulation Function (ANSWERS) Eample. Find a function y=f() whose derivative is that f()=. dy d tan that satisfies the condition We can use the Fundamental Theorem to write a function y f( ) tand whose derivative is tan :. Enter this equation in y. And graph it in a window: [-.5,.5,,-4.,4.,] Make a sketch in the graph at the right. Notice that this function does not satisfy the condition that f()=. In order to meet the condition that f()= we can make a vertical shift. Adjust the equation in y with a vertical shift of. Write the equation that will shift the graph vertically units. y tan( tdt ) This new function now passes through (,). Make a sketch of your new shifted graph. Why is the graph of y the given shape? Add the graph of y = tan. Superimpose this on the other graph. Since tan is negative when -.5 < < the graph of y is decreasing. Since tan is positive when < <.5 the graph of y is increasing. There is a relative minimum at = since tan changes from negative to positive at =. If y f( ) tand shape., eplain why the graph of y has the particular Rahn (c) 3

2 Consider the following regions for : = : y. y f() tand This shows the y-intercept for the graph of.5 : When is in this region tan d will be positive since we are integrating in the reverse direction. Since the graph of y= ta is below the -ais, and we are integrating in reverse, the graph of y will be positive. The largest value for y will occur at -.5 and the smallest value at =..5: When is in this region tan d will be positive since we are integrating in the normal direction. Since the graph of y= ta is above the -ais, and we are integrating in the normal direction, the graph of y will be positive. The largest value for y will occur at.5 and the smallest value at =. dy dy If tan is the derivative of y, then the tan describes how the d d graph of y behaves. dy Using the graph of tan decide the intervals where y=f() is increasing and d decreasing. Confirm that y does increase and decrease in the correct locations. dy Since tan is negative when.5, the graph of y is decreasing d dy in that region. Since tan is positive when.5, the graph of y is d increasing in that region. The graph of y has a minimum at = because the dy graph of tan changes from being negative to positive at =. d dy Use the graph of tan to decide where y=f() is concave up and concave d down. Confirm that y does have the correct concavity. dy Since the slope of the graph of tan d of y is concave up ecept at =. greater than or equal to zero, the graph Rahn (c) 3

3 Eample : 4, Let f( )., The function is illustrated in the following graph. If this function (or graph) represents f() and F ( ) ftdt ( ). Find F() f( t) dt Find F() f( t) dt radius of. since this definite integral is a quarter of a circle with a Find F(3) f( t) dt ()() 3 Try to picture what the graph of F would look like. Think of some addition values of F() that would help you create a graph. With the help of a TI-84 you can create a graph of this new function F() without having to know the antiderivative. The calculator can approimate the definite integrals. Enter the equations. y3 4t dt y4 dt Make a sketch of F(). When is F() negative? Why is this? F() is negative when. This is because we are integrating in the reverse direction and the region is above the - ais. Rahn (c) 3

4 When is F() positive? Why is this?? F() is positive when. This is because we are integrating in the in the positive direction and the region is above the -ais. Using the Accumulation Function with Composite Functions a F(g()) = Suppose instead of F() = f(t)dt we had. How would you write F'(g())? g() a f(t)dt d g() f(t)dt a F'(g()) = f( g( )) g'( ) d Try this AP Problem: Let F() = t tdt A. Find F (). F'() = t tdt B. Find the domain of F: lim F ( ) C. Find : t or t lim F( ) lim t tdt t tdt Rahn (c) 3

5 A Second AP Problem The graph of the differentiable function f on the closed interval [,7] is shown at the right. h ( ) ftdt ( ) 7 Let for. A. Find h() B. Find h'(4) h() f( t) dt h'( ) f( ) h'(4) f(4).8 C. On what interval or intervals is the graph of h concave upward? Justify your h'( ) f( ) h"( ) f'( ) Graph of f answer. So h ()> when f ()> or or D. Find the value of at which h has its minimum on the closed interval [,7]. Justify your answer. Possibilities for the location of minimum values would be at the endpoints or any critical values. The only critical value for h is at =5.. h() h(5.) h(7) h(5.) So the minimum value occurs at =. Rahn (c) 3

6 A Third AP Problem y Let f be a function whose domain is the closed interval [,5]. The graph of f is shown at the right. 3 h ( ) ftdt ( ) Let. A. Find the domain of h. B. Find h'(): h'( ) f 3 3 h'() f 3 f 4 3 C. At what is h() minimum? Show the analysis that leads to your conclusion. Graph of f To determine where h has a minimum we need to study h (). h'( ) f 3 6 The inside function will change the domain from to because of the. Then the domain will be shifted 3 9 three units to the left or because of the 3. The sign of h changes from positive to negative only once at =. Therefore this is the location of the minimum value. Rahn (c) 3