# Have no idea what to do..

• Jan 4th 2009, 09:57 AM
elpermic
Have no idea what to do..
If the function f(x) is differentiable and

f(x)=
{ax^3 + 6x, if x≤1
{ $bx^2 + 4, if x>1$

then a =

What do I do?? No idea what's going on..
• Jan 4th 2009, 10:02 AM
Chris L T521
Quote:

Originally Posted by elpermic
If the function f(x) is differentiable and

f(x)=
{ax^3 + 6x, if x≤1
{ $bx^2 + 4, if x>1$

then a =

What do I do?? No idea what's going on..

You need to do two things.

First, in order for it to be differentiable, it must be continuous at that point.

So you want to see where $\lim_{x\to1^-}f\left(x\right)=\lim_{x\to1^+}f\left(x\right)$.

Then. For it to be differentiable at that point, the slopes of the two piecewise functions must be the same at that point. This means that $3a+6=2b$

You will then be able to come up with a system of equations. Now you can solve it for a.

Does this make sense? Can you take it from here?
• Jan 4th 2009, 11:26 AM
elpermic
I got my answer as b=4, a=2

I don't think it is right however..
• Jan 4th 2009, 11:47 AM
Chris L T521
Quote:

Originally Posted by elpermic
I got my answer as b=4, a=2

I don't think it is right however..

I got a=-2 and b=0. (Thinking)

What did you get for your system of equations?
• Jan 4th 2009, 11:48 AM
skeeter
Quote:

Originally Posted by elpermic
I got my answer as b=4, a=2

I don't think it is right however..

why don't you?
• Jan 4th 2009, 12:01 PM
elpermic
a-b=-2
a-2b=-6
• Jan 4th 2009, 12:01 PM
elpermic
Quote:

Originally Posted by skeeter
why don't you?

These were the choicces..

a)0
b)1
c)-14
d)-24
e)26
• Jan 4th 2009, 12:33 PM
skeeter
I get a = -2 also ... recheck the original function, make sure you copied it down correctly.
• Jan 4th 2009, 01:31 PM
elpermic
Double checked, it is the same exact function as it says on my paper..

Don't know what's up with this one..