Find A and B so that F(x) is a Differentiable Function

[Jakey214]

Homework Statement

Find the values of a and b that make f a differentiable function.

Note: F(x) is a piecewise function

f(x):
Ax^2 - Bx, X ≤ 1
Alnx + B, X > 1

Homework Equations

The Attempt at a Solution

Made the two equations equal each other.
Ax^2 - Bx = Alnx + B
Inserting x=1 gives,
A - B = B, which is also A - 2B = 0, which also means A = 2B

Deriving the equation,
2Ax - B = A/X
Inserting x=1 here gives,
2A - B = A, which is also A - B = 0, which also means A = B

By then I'm stumped here.
I try to eliminate either A and B with,
A - B = 0
A - 2B = 0
In the end, both A and B would have to equal zero, both of which doesn't work.

If I have A = 2B then,
F'(x): A - B = 0 ⇒ 2B - B = 0 ⇒ B = 0
As said before, B = 0 would not be the right answer, as far as I know at least.

If A = B, then
F(x): A - 2B = 0 ⇒ B - 2B = 0 ⇒ B = 0
Once Again, B equaling zero would not work.

Right now, I'm convinced this problem is virtually impossible.

 

[PetSounds]

Are you sure you copied the problem down correctly?

 

[Jakey214]

Are you sure you copied the problem down correctly?

Yep, that's exactly what it says.

 

[PetSounds]

Yep, that's exactly what it says.

Then I'm stumped too. Perhaps someone else will have a solution.

 

[Jakey214]

Probably should have put this in earlier as proof of the existence of the problem.

Hopefully links are fine. Sorry if it isn't allowed.

http://www40.zippyshare.com/i/3Pmk4c2p/82894/WIN_20170301_215129.JPG

 

[LCKurtz]

So what is wrong with $f(x) \equiv 0$? That's a perfectly good differentiable function.

 

[Jakey214]

UPDATE:

My Math Teacher just said it was a typo, where he forgot to add a negative two, so that the piecewise function would be:

f(x):
Ax^2 - Bx - 2, X ≤ 1
Alnx + B, X > 1

Now solving it, I got A = -2 and B = -2.