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

[FritoTaco]

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?

 

[Mark44]

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?

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

 

[PeroK]

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 $3+1$?

 

[FritoTaco]

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.

 

[PeroK]

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.

 

[BvU]

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 $$=3a^2-3b+3 = 9a - 3b + 3 \ (4)$$

 

[FritoTaco]

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?

 

[BvU]

Wait, soo. Is this okay?

No it is not. How can you ask such a thing ?$$\lim_{x\downarrow 2} \;ax^2-bx+3 = a\;2^2 - b\;2 + 3 = 4a-2b+3$$

 

[FritoTaco]

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 ax^2-bx+3?

 

[BvU]

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 $$
\quad \quad \quad\quad \quad \quad \quad \quad \quad = 9a-3b+3 \quad\quad (4) $$

 

[FritoTaco]

Oh. Sorry again ಠ_ಠ (got my serious face on). Okay, so for the \lim\limits_{x \to 2}ax^2-bx+3 = 2a^2-2b+3 = 4a-2b+3 I don't see how that's wrong. 2^2 = 4 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 2a^2, a is being squared. It could've been a2^2 instead, but for the sake of looking nice, I had it the other way around.

 

[BvU]

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 $a2^2$.

 

[Mark44]

Making things look nice can get you into trouble.

It's better to write something correct and have it look so-so, than to write some incorrect and have it look elegant.

 

[FritoTaco]

Guys, I'm talking about how it's like this: ax^2, so it's a2^2. Then from a4, I switched it around to 4a. To my knowledge, a4 and 4a are equal in sense.

 

[SammyS]

You just made a simple error getting to equation 6. Hint: what is $3+1$?

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).

I'm pretty sure @PeroK was referring to the following step

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 $a$.

 

[BvU]

But of course $3a^2$ looks a lot nicer than $a3^2$ .

 

[FritoTaco]

Alright, after sleeping I've come to my senses. Sorry about that. So I got:

9a-3b+3=6-a+b (4=5)

4a-2b=1 (3)
10a-4b=3 (6)

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

2a=1

a=\dfrac{1}{2}

Plug a in for (3)

4(\dfrac{1}{2})-2b=1

2-2b=1

b=\dfrac{1}{2}

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. :)