Question

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show all the work ( explain)
do not use a calculator for parts 1-4.
Let f be the function defined as follows:

Answer 1

1. if a=2 and b=3, is f continuous at x=1? justify your answer.
2. find a relationship between a and b for which f is continuous at x=1. ( Hint: A relationship between a and b just means an equation in a and b.
3. Find a relationship between a and b so that f is continuous at x=2.
4. USe your equations from parts (ii)and(III) to find the values of a and b so that f is continuous at both x=1 and at x=2?
5. Graph the piece function using the values of a and b that you have found. you may graph by hand or use your calculator to graph.

Answer 2

Any idea?

Answer 3

no im completely clueless in this! o.o

Answer 4

1) No, since we don't have x =1 on f(x). (The way they define f(x) excludes 1)
2) if f(x) is continuous at x =1, that is the first part = second part (in f(x)
I meant if x =1, then y =2 (for f(x) =3-x)
Plug it into the second one ax^2 + bx =y with x =1, y =2 to have the relationship between a and b
3) do the same with the second part and the third part for x =2
That is if x =2, y = 0 (for y = 5x-10)
Now we have (2,0) , replace in ax^2+bx =y to get the relationship between them.

Answer 5

4) combine part 2 and part 3 to solve for a, b. We have 2 equations, 2 unknowns --> the system is solvable.
5) graph, this part is the easiest part. I think you can do it, right?

Answer 6

huuuuuuhhhhh? o.ohow did you get that?

Answer 7

You asked "how did I get that?" , right? of course, I got it from the courses I took in the past.

Answer 8

lol I mean how did you arrive to the answers?

Answer 9

I don't get what you mean. My logic is all in what I said. Read and you can see how, why and answer.

Answer 10

ok if I need any more help can I message you

Answer 11

you got this?

Answer 12

is there anything you wold like to add?

Answer 13

i was just wondering if you got the answer
the question is just walking you through it step by step

Answer 14

not really im tryig on my own lol. but I am terrible. even with the steps which I appreciate , I am having a little trouble getting to the answer.

Answer 15

this is what you do
plug in 1 for \(x\) in both \(3-x\) and \(ax^2+bx\) you get
\[3-1=2\] and
\[a+b\] setting them equal tells you
\[a+b=2\]

Answer 16

then plug in \(2\) in \(ax^2+bx\) and \(5x-10\)
you get
\[4a+2b\] and \[2\times 2-1=0\] telling you
\[4a+b=0\]

Answer 17

is this for number one?

Answer 18

now you have
\[a+b=2\\
4a+2b=0\] solve that for \(a\) and \(b\) and you are done

Answer 19

wait do I plug in a+b=2 into 4a+2b=0?

Answer 20

you solve the "system of equations" just as normal
\[a+b=2\\
4a+2b=0\] you can solve it using any method you like (or know)

Answer 21

ok this is what I get:

4a^2+ 2b^2=2