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])
i love seeing all the answers from the 2000s makes me feel nostalgic
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.
do u still have a test tomorrow
To solve these questions, we need to understand the concept of the definite integral and utilize the properties of the graph given. Here's how you can solve each part:
(a) To find g(1), we need to evaluate the definite integral of f(t) from 1 to 1. Since the upper and lower limits of integration are the same, this integral evaluates to 0.
(b) To find g(3), we need to evaluate the definite integral of f(t) from 1 to 3. By carefully analyzing the graph, we can see that this integral consists of the semi-circle and the two line segments. The semi-circle has a negative area, while the line segments have positive areas. Therefore, the sum of these areas will result in a negative value. Hence, g(3) = -1.
(c) To find g(-1), we need to evaluate the definite integral of f(t) from 1 to -1. Similar to part (b), we can observe that the semi-circle has a negative area, while the line segments have positive areas. The semi-circle's area can be calculated as -pi. Adding the areas of the line segments to this value will give us the result. Therefore, g(-1) = -pi.
(d) To find the values of x on the open interval (-3,4) at which g has a relative maximum, we need to look for the points on the graph where the slope changes from positive to negative or vice versa. From the given graph, we can see that there is only one such point, which is x = 1. Hence, x = 1 is the only value on the interval at which g has a relative maximum.
(e) To write an equation for the line tangent to the graph of g at x = -1, we can find the derivative of g(x) with respect to x. By doing this, we get g'(x) = f(x). Now, substitute x = -1 into g'(x) to find the slope of the tangent line at x = -1. Plug this slope and the point (-1, g(-1)) into the point-slope form of a linear equation to obtain the equation of the tangent line, which is y = 2x + 2pi.
(f) To find the x-coordinate(s) of each point of inflection of the graph of g on the open interval (-3,4), we need to look for points where the concavity changes. In this case, we can see that the concavity changes at x = -1 and x = 2. Thus, x = -1 and x = 2 are the x-coordinates of the points of inflection.
(g) To find the range of g, we need to consider the y-values (outputs) of g(x). From the given graph, we can see that the minimum value of g(x) occurs when the semi-circle is at its lowest point, which is -2pi. The maximum value of g(x) can be found at the top of the semi-circle, which is 0. Therefore, the range of g is [-2pi, 0].
Remember, it's crucial to understand the concepts and properties involved to solve these types of problems effectively. Good luck with your test!