Ind the Values of a and B That Make F Continuous Everywhere
Find the values of a and b that make f continuous everywhere
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Homework Statement
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.
Homework Equations
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?
Answers and Replies
For the record, here's the problem description:Homework Statement
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.
Homework Equations
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?
Find the values of a and b that make f continuous everywhere.
$$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x < 2 \\ ax^2 - bx + 3 & 2 \le x < 3\\ 2x - a + b& x \ge 3\end{cases}$$
Homework Statement
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.
Homework Equations
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?
You just made a simple error getting to equation 6. Hint: what is ##3+1##?
You just made a simple error getting to equation 6. Hint: what is ##3+1##?
It's 4. But why am I adding the constant terms from (3) and (6)? I was trying to first get b cancelled.
It's 4. But why am I adding the constant terms from (3) and (6)? I was trying to first get b cancelled.
You did ##3 + 1 = 5##. If you do ##3+1 =4## the problem can be solved.
I shudder when I read $$(2) =4a - 2b + 3 $$right below $$= 2a^2-2b+3$$ and when I read $$=3a^2-3b+3 = 9a - 3b + 3 \ (4)$$
You did ##3 + 1 = 5##. If you do ##3+1 =4## the problem can be solved.
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).
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 $$=3a^2-3b+3 = 9a - 3b + 3 \ (4)$$
Wait, soo. Is this okay?
No it is not. How can you ask such a thing ?$$\lim_{x\downarrow 2} \;ax^2-bx+3 = a\;2^2 - b\;2 + 3 = 4a-2b+3$$
No it is not. How can you ask such a thing ?$$\lim_{x\downarrow 2} \;ax^2-bx+3 = a\;2^2 - b\;2 + 3 = 4a-2b+3$$
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]?
\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) $$
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.
Look at it again tomorrow. ##2a^2## may look nicer, but it is NOT the same as ##a2^2##.
It's better to write something correct and have it look so-so, than to write some incorrect and have it look elegant.Making things look nice can get you into trouble.
You just made a simple error getting to equation 6. Hint: what is ##3+1##?
I'm pretty sure @PeroK was referring to the following stepOh, 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).
That should not be -5, Right ?
As far as the comments regarding careless/sloppy notation:
You have
but it's the 3 that's to be squared, not the ##a##.
.[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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