Problem set 2 2.2.1. section 2.1. (1) let l be the operator defined by l() = (á - a) a. show that l is linear. b. show

5 answers
Question:

Problem set 2 2.2.1. section 2.1. (1) let l be the operator defined by l() = (á - a) a. show that l is linear. b. show that the kernel of l is a vector space. c. let g(t) be a non-trivial function. show that the collection of vector-valued functions which satisfy = g(t) do not form a vector space. d. show that the difference between any two functions, t1, t2 which satisfy l(t) = g(t) is in the kernel of l. ) read the statement of abel's theorem. a. what are the hypotheses of abel's theorem as stated in the notes? b. what are the conclusions of abel's theorem as stated in the notes? c. show that ? i(t) = (+) and tz(t) = (2) are solutions to = ( ) [e). d. what are the largest intervals on which õi(t) and 72(t) are linearly indepen- dent? e. does this violate abel's theorem as stated in the notes? why or why not?

Answers

1)

Option: C is correct.

( The car travels different distances per unit of time because the portion shows a nonlinear, increasing function ).

As the graph is a curve so the distance traveled is different at different times also it is given that the curve is up.

2)

The graph below plots a function f(x):

graph of line segment going through ordered pairs (0,150) and (3, 0 ).

The rate of change of the function f(x) is given as:

[tex]\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{0-150}{3}\\\\=-50[/tex]

The average rate of change of the function in first three seconds is: -50.

3)

Table A

x332

y100

Table B

x355

y−22−2

The correct statement is:

Both Table A and Table B do not represent functions.

( Because a function is a relation such that every element has a unique image.

In table A:  3 has two images or we can say y-value (1 and 0)

similarly in Table B:  5 has two values (2 and -2) )

4)

What is the slope of the line that passes through (3, −7) and (−1, 1)?

We know that the equation of the line in slope intercept form is given as:

y=mx+c where m represents the slope and c represents the y-intercept.

The equation of line passing through two points (a,b) and (c,d) is given by:

[tex]y-b=\dfrac{d-b}{b-a}\times (x-a)[/tex]

Here we have (a,b)=(3,-7) and (c,d)=(-1,1)

hence equation of line is:

[tex]y-(-7)=\dfrac{1-(-7)}{-1-3}\times (x-3)\\\\y+7=\dfrac{8}{-4}\times (x-3)\\\\y+7=-2\times (x-3)\\\\y+7=-2x+6\\\\y=-2x+6-7\\\\y=-2x-1[/tex]

Hence the slope of line is: -2.

5)

The domain of the graph are the possible values where the function is defined.

Now here we are given graph of a relation as:

Discrete graph defined on the set of points negative two comma negative four and zero comma zero and positive one comma positive one and positive two comma positive two and positive three comma positive one and positive four comma positive six.

in table form we can write it as:

x            -2     0    1     2     3    4

y             -4     0    1     2     1     6

Hence the domain is the set of x-values.

i.e. domain is:

{-2,0,1,2,3,4}.

A 7 and 5 differences are 4 and -2

b) 4, K

c)-2

d) no depends on K

I have no answer for the second part yet

Step-by-step explanation:

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Step-by-step explanation:

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The car travels different distances per unit of time because the portion shows a nonlinear, increasing function; -50; Both Table A and Table B do not represent functions; −2 ; {−2, 0, 1, 2, 3, 4}

Step-by-step explanation:

A linear graph of any sort is a graph that has a constant rate of change.  This graph is not linear; therefore it does not have a constant rate of change.  This means that the car travels different distances per unit of time between sections.

To find the rate of change, we find the slope between these two points.  We use the formula

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Using our points, we have

[tex]m=\frac{150-0}{0-3}=\frac{150}{-3}=-50[/tex]

A function is a relation in which each element of the domain (x) is mapped to only one element of the range (y).  In both of these tables, we have x-values that are mapped to two y-values; in Table A, 3 is mapped to 1 and 0; and in Table B, 5 is mapped to 2 and -2.

We use the slope formula again:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Using our points, we have

[tex]m=\frac{-7-1}{3--1}=\frac{-7-1}{3+1}=\frac{-8}{4}=-2[/tex]

The domain of a graph is the set of inputs or x-values.  From our points, we have the set:

{-2, 0, 1, 2, 3, 4}

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