This 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.
Help! I have a test tommorow! I don't understand (b), (c), (e), and (g). The answers are listed following the each question. Here's a discription of the graph:
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])
3 answers
do u still have a test tomorrow
i love seeing all the answers from the 2000s makes me feel nostalgic
3 answers