Question

I NEEEEED HHHHHEEEEELLLLLPPPP PLEASE!!!!

Answer 1

For what values of a and b that make the function f continuous everywhere.
http://roy.math.umn.edu/webwork2_files/tmp/equations/26/eacd2b1d747d507691a8d8a0502ec81.png
a= b=

Answer 2

@Directrix

Answer 3

@Callisto

Answer 4

We are studying this right now in my calculus class so I won't answer as I'm not that confident and don't want to tell you something wrong but for these I find it best to graph all of the functions and find between what two points will it be continuous.

Answer 5

what do i do about the a's and b's

Answer 6

I think that a is your x-value on the left and b is your x-value to the right. It is between these two values that it is continuous.

Answer 7

so what do i do then

Answer 8

Graph all of the equations and then look at the graph to see where it is continuous for all of them.

Answer 9

how do i graph them with the a's and b's

Answer 10

Oh I see what you mean... like I said we just started doing this but maybe try factoring the second one and I think it will give you x-intercepts?? What I mean is put it in factored form, the third one I'm not sure of... sorry

Answer 11

no its ok

Answer 12

@Directrix @jim_thompson5910

Answer 13

http://www.physicsforums.com/showthread.php?t=259382 I haven't finished reading it myself but give it a try.

Answer 14

thanks

Answer 15

You're welcome. Good luck

Answer 16

you should choose a and b so that the border values in each of the three cases is same

Answer 17

2 or 3

Answer 18

all the 3 cases

Answer 19

take first equation and tell me what is the value of expression when x->2

Answer 20

4

Answer 21

correct..........so when x=2 in the second equation, it should also give 4 ------so that the function is continuous...right?

Answer 22

yup

Answer 23

so when i put x=2 in the second equation - what do you get?

Answer 24

you will get something like this
4a - 2b -20 =4

Answer 25

yup

Answer 26

similarly do this with the border values of 2nd and 3rd equation when x =3.......you will get one more equation - you can solve them both to get the values of a and b

Answer 27

yuo get two equations 9a-3b-20 and 36-a-b @atlas

Answer 28

yeah and they must be equal to have continuous value at 3..right??
so 9a-3b-20 = 36-a-b

Answer 29

\[\lim_{x \rightarrow 2-}\frac{ x ^{2} -4}{ x-2 }=\lim_{x \rightarrow 2-}\frac{ \left( x+ 2 \right)\left( x-2 \right) } { x-2 } =\lim_{x \rightarrow 2- } \left( x+2 \right)\]
[put x=2-h,h>0 \[h \rightarrow 0 as x \rightarrow2-]\]
\[=\lim_{h \rightarrow0}\left( 2-h+2 \right)=2-0+2=4 .....(1)\]

Answer 30

ok so i have my two equations 4a-2b-24 and 10a-4b-56

Answer 31

You have two equations with you now and two unknown variables:
4a -2b-20 =4 (which we got earlier)
and 9a -3b-20 = 36-a-b (which we got now)
I guess you know how to solve them to get the values of a and b

Answer 32

you got the 2nd equation wrong ----------it is noe 10a -'4'b -56.........check again

Answer 33

*not

Answer 34

10a-2b-56

Answer 35

right!

Answer 36

now solve the two equations and get the answer

Answer 37

so i just solve for a and b

Answer 38

yup and you have your answer :P

Answer 39

i got to go now......have a good time at OS

Answer 40

4a-2b-24 =0
10a -2b-56 =0
Subtract equation i from ii

Answer 41

\[\lim_{x \rightarrow 2+}\left( ax ^{2} -bx-20\right)\]put x=2+h
\[h \rightarrow 0 asx \rightarrow 2+\]
\[\lim_{h \rightarrow0 }\left\{ a \left( 2+h \right)^{2}+b \left( 2+h \right)-20 \right\}\]
\[=a \left( 2+0 \right)^{2}+b \left( 2+0 \right)-20=4a+2b-20 .....(2)\]
because f(x) is continuous ,hence from (1) and (2)
4a+2b-20=4
or 2a+b-12=0 ....(3)

Answer 42

the b's cancel out though

Answer 43

yeah thats ok.....find a

Answer 44

i got 16/3 and it says its wrong

Answer 45

then use the value of a in any equation to get b......hope that helps

Answer 46

k equation should be 4a + 2b -24 =0...........check again and solve it

Answer 47

it should be +2b and not -2b......that will correct your answer

Answer 48

\[\lim_{x \rightarrow 2+}\left( ax ^{2} -bx-20\right)\]put x=2+h
\[h \rightarrow 0 asx \rightarrow 2+\]
\[\lim_{h \rightarrow0 }\left\{ a \left( 2+h \right)^{2}+b \left( 2+h \right)-20 \right\}\]
\[=a \left( 2+0 \right)^{2}+b \left( 2+0 \right)-20=4a+2b-20 .....(2)\]
because f(x) is continuous ,hence from (1) and (2)
4a+2b-20=4
or 2a+b-12=0 ....(3)