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Find the values of a and b that make f continuous everywhere

  1. Jan 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the values of a and b that make f continuous everywhere.

    See attachment for the function.

    I'm suppose to find a and b.

    2. Relevant equations

    3. The attempt at a solution

    See the second attachment

    The problem I have is, when I get to the last step, I'm trying to cancel out b so I can get just a. When I do that by manipulating the equation (multiplying (3) and (6) the orange writing you see on the board) it ends up canceling out a (so both a and b). Did I go wrong somewhere?
     

    Attached Files:

  2. jcsd
  3. Jan 26, 2017 #2

    Mark44

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    For the record, here's the problem description:
    Find the values of a and b that make f continuous everywhere.
    $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x < 2 \\ ax^2 - bx + 3 & 2 \le x < 3\\ 2x - a + b& x \ge 3\end{cases}$$
     
  4. Jan 26, 2017 #3

    PeroK

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    You just made a simple error getting to equation 6. Hint: what is ##3+1##?
     
  5. Jan 26, 2017 #4
    It's 4. But why am I adding the constant terms from (3) and (6)? I was trying to first get b cancelled.
     
  6. Jan 26, 2017 #5

    PeroK

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    You did ##3 + 1 = 5##. If you do ##3+1 =4## the problem can be solved.
     
  7. Jan 26, 2017 #6

    BvU

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    The answer to your question is yes.
    I shudder when I read $$(2) =4a - 2b + 3 $$right below $$= 2a^2-2b+3$$ and when I read $$=3a^2-3b+3 = 9a - 3b + 3 \ (4)$$
     
  8. Jan 26, 2017 #7
    Oh, no I wasn't solving (3) and (6) by adding/subtracting. I was manipulating both equations by finding the LCD for b so I can cancel the terms out, and when I do that I multiplied the whole equation. As well as equation (6).

    Wait, soo. Is this okay?
     
  9. Jan 26, 2017 #8

    BvU

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    No it is not. How can you ask such a thing ?$$\lim_{x\downarrow 2} \;ax^2-bx+3 = a\;2^2 - b\;2 + 3 = 4a-2b+3$$
     
  10. Jan 26, 2017 #9
    Sorry lol. Okay, but I thought I am looking at limit as x approaches 3. So won't I plug in 3 into [itex]ax^2-bx+3[/itex]?
     
  11. Jan 26, 2017 #10

    BvU

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    lol ? It's a crying matter ! On the left I clearly read $$
    \lim_{x\downarrow 2} \;ax^2-bx+3 =2 a^2-2 b+3 $$ followed by $$
    \quad \quad \quad \quad \quad (2) = 4a-2b+3$$

    And on the right side it happens again ! $$
    \lim_{x\uparrow 3} \;ax^2-bx+3 =3 a^2-3b +3 $$ followed by $$
    \quad \quad \quad\quad \quad \quad \quad \quad \quad = 9a-3b+3 \quad\quad (4) $$
     
  12. Jan 26, 2017 #11
    Oh. Sorry again ಠ_ಠ (got my serious face on). Okay, so for the [itex]\lim\limits_{x \to 2}ax^2-bx+3 = 2a^2-2b+3 = 4a-2b+3[/itex] I don't see how that's wrong. [itex] 2^2 = 4[/itex] if i'm correct, and so forth. Am I going crazy?

    Edit: Unless it's the notation you're looking at where the way I had it [itex]2a^2[/itex], a is being squared. It could've been [itex]a2^2[/itex] instead, but for the sake of looking nice, I had it the other way around.
     
    Last edited: Jan 26, 2017
  13. Jan 26, 2017 #12

    BvU

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    I give up. Making things look nice can get you into trouble.
    Look at it again tomorrow. ##2a^2## may look nicer, but it is NOT the same as ##a2^2##.
     
  14. Jan 26, 2017 #13

    Mark44

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    It's better to write something correct and have it look so-so, than to write some incorrect and have it look elegant.
     
  15. Jan 26, 2017 #14
    Guys, I'm talking about how it's like this: [itex]ax^2[/itex], so it's [itex]a2^2[/itex]. Then from [itex]a4[/itex], I switched it around to [itex]4a[/itex]. To my knowledge, [itex]a4[/itex] and [itex]4a[/itex] are equal in sense.
     
  16. Jan 26, 2017 #15

    SammyS

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    I'm pretty sure @PeroK was referring to the following step
    upload_2017-1-26_19-14-10.png
    That should not be -5, Right ?

    As far as the comments regarding careless/sloppy notation:
    You have
    upload_2017-1-26_19-17-35.png ,
    but it's the 3 that's to be squared, not the ##a##.
     
  17. Jan 27, 2017 #16

    BvU

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    But of course ##3a^2## looks a lot nicer than ##a3^2## :smile:.
     
  18. Jan 27, 2017 #17
    Alright, after sleeping I've come to my senses. Sorry about that. So I got:

    [itex]9a-3b+3=6-a+b[/itex] (4=5)

    [itex]4a-2b=1[/itex] (3)
    [itex]10a-4b=3[/itex] (6)

    [itex]-8a+4b=-2[/itex] (3). Multiply equation by -2
    [itex]10a-4b=3[/itex] (6)

    [itex]2a=1[/itex]

    [itex]a=\dfrac{1}{2}[/itex]

    Plug a in for (3)

    [itex]4(\dfrac{1}{2})-2b=1[/itex]

    [itex]2-2b=1[/itex]

    [itex]b=\dfrac{1}{2}[/itex]

    As far as notation, from my eye, I knew what it should've looked like and I didn't write it out like it should've been, but I'll be careful next time on how I should write it.

    Thank you guys. :)
     
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