# Thread: continuity and differentiablity question again.

1. ## continuity and differentiablity question again.

A function f(x) is defined to equal cos(ax) + b for x>=0 and 2 – x3 for x < 0, where a and b are constants. It is known that this function is differentiable everywhere. Find all possible values of a and b

I posted this quesiton earlier, and I performed the steps as you indicated, ie lim x-->0+ f'(x) = lim x-->0- f'(x), so
lim x-->0- f'(x) = -3x^2 = lim x-->0+f'(x) = -asin(ax), and after some more math, I ended up with 0 = 0, which I concluded that a could be anything. I know theere is something wrong. This is why I posted this question.

Thank you.

2. I don't see this as being possible. If I'm reading this right, what needs to happen is that both parts of the function need to have the same slope at x=0. So:

$\displaystyle \frac{d}{dx}(\cos{ax}+b) = \frac{d}{dx}(2-3x)$

$\displaystyle -a\sin{ax}=-3$

$\displaystyle -a\sin{0}=-3$

$\displaystyle 0=-3$

And obviously that can't be. So, I don't think it is possible for f(x) to be differentiable for all x.

3. ## continuity and differentiability

the second function is f(x) = 2-x^3, not f(x)= 2-3x

4. Originally Posted by geometry101
A function f(x) is defined to equal cos(ax) + b for x>=0 and 2 – x3 for x < 0, where a and b are constants. It is known that this function is differentiable everywhere. Find all possible values of a and b

I posted this quesiton earlier, and I performed the steps as you indicated, ie lim x-->0+ f'(x) = lim x-->0- f'(x), so
lim x-->0- f'(x) = -3x^2 = lim x-->0+f'(x) = -asin(ax), and after some more math, I ended up with 0 = 0, which I concluded that a could be anything. I know theere is something wrong. This is why I posted this question.

Thank you.
This is exactly the same as the other problem I answered, which you never responded to. Do you understand that?

5. ## in response to mathstud

Yes, I followed what you had stated. THis was my originial solution, and what I stated was that I arrived at 0 = 0, hence I concluded that a could be any value. Work it for yourself, and you will see this is true.