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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.
no im completely clueless in this! o.o
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.
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?
huuuuuuhhhhh? o.ohow did you get that?
You asked "how did I get that?" , right? of course, I got it from the courses I took in the past.
lol I mean how did you arrive to the answers?
I don't get what you mean. My logic is all in what I said. Read and you can see how, why and answer.
ok if I need any more help can I message you
you got this?
is there anything you wold like to add?
i was just wondering if you got the answer the question is just walking you through it step by step
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.
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 \[a+b=2\]
then plug in \(2\) in \(ax^2+bx\) and \(5x-10\) you get \[4a+2b\] and \[2\times 2-1=0\] telling you \[4a+b=0\]
is this for number one?
now you have \[a+b=2\\ 4a+2b=0\] solve that for \(a\) and \(b\) and you are done
wait do I plug in a+b=2 into 4a+2b=0?
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)
ok this is what I get: 4a^2+ 2b^2=2