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August 28, 2014

August 28, 2014

Posted by **Anonymous** on Friday, October 26, 2007 at 12:31am.

There is a graph of a function f consists of a semi circle (-3 to 1 faced downward on the x-axis) and two line segments. One segment is from (1,-1) to (2-1), and another segment is from (2,-1) to (3,1).

Let g(x)=S(1 to x)f(t)dt

(a) Find g(1)=0

(b) Find g(3)=-1

(c) Find g(-1)=-pi

(d) Find all values of x on the open interval (-3,4) at which g has a relative maximum. (answer:x=1)

(e) Write an equation for the line tangent to the graph of g at x=-1.

(answer:y=2x+2pi)

(f) Find the x-coordinate of each point of inflection of graph of g on the open interval (-3,4).

(answer:x=-1, x=2)

(g) Find the range of g.

(answer:[-2pi,0])

- calculus -
**drwls**, Friday, October 26, 2007 at 11:48amThis problem can be done almost by inspection if you just draw a graph of the function and think about what each question means in terms of the graph. For (a), g(x) is the area of a curve that starts AND ends at x=1. That area is obviously zero. For (b), the area of the portion from 1 to 3 is

-1 x 1 (from x = 1 to 2),

-(1/4) (2 < x < 2.5)

+ (1/4) (2.5 < x < 3.

The sum of the areas is -1. . Try your hand at the rest. We will be happy to critique your reasoning and results.

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