Mathematics, 16.07.2019 07:30 tinajackson6534

Question 1(multiple choice worth 2 points) use the given graph to determine the limit, if it exists. a coordinate graph is shown with a downward sloped line crossing the y axis at two that ends at the open point 2, 1.3, a closed point at 2, -1, and an upward sloped line starting at the open point 2, 4. find = limit as x approaches two from the right of f of x.. 1.3 4 -1 5 question 2(multiple choice worth 4 points) use the given graph to determine the limit, if it exists. a coordinate graph is shown with a horizontal line crossing the y axis at three that ends at the open point 2, 3, a closed point at 2, 1, and another horizontal line starting at the open point 2, -3. find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x.. 1; 1 3; -3 does not exist; does not exist -3; 3 question 3 (essay worth 4 points) use the given graph to determine the limit, if it exists. a coordinate graph is shown with a downward sloped line crossing the y axis at the origin that ends at the open point 3, -1, a closed point at 3, 7, and a horizontal line starting at the open point 3, -4. find limit as x approaches three from the left of f of x..

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These are 3 questions and 3 answers.

1) Find

$\lim_{x \to \ 2^+} f(x)$

4.

Explanation:

The expression means the limit as the function f(x) approaches 2 from the right.

Then, you have to use the function (the line) that comes from the right of 2 and gets as close as you want to x = 2.

That is the line that has the open circle around y = 4, and that is the limit searched.

2) Use the graph to determine the limit if it exists.

$\lim_{x \to \ 2^-} f(x) = 3$

$\lim_{x \to \ 2^+} f(x)=-3$

To determine each limit you use the function from the side the value of x is being approached.

Note, that since the two limits are different it is said that the limit of the function as it approaches 2 does not exist.

3)
- 1

$\lim_{x \to \ 3^-} f(x) = -1$

To find the limit when the function is approached to 3 from the left you follow the line that ends with the open circle at (3, -1).

Therefore, the limit is - 1.

Answer from: Quest

Sorry can u repeat that?

Answer from: Quest

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 = (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4 we need to find some relationship between a and b.

let's try ab = bc

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2

br = sqrt(3) * a

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram.

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a

cp = sqrt(3)*a

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt 3) * a which agrees with the other two.

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