# Continuity problem

• Jan 30th 2013, 09:59 PM
Paze
Continuity problem
Find values of the constants a and b for which the following function is continuous but not differentiable.

$\displaystyle f(x)=\left\{\begin{array}{cc}ax+b,&\mbox{ if }x>0;\\sin2x, & \mbox{ if } x\leq 0\end{array}\right.$

In other words, the graph of the function should have a sharp corner at the pont (0,f(0)).

I'm not even sure what they are asking. This is part of a limit course and I am very familiar with limits.
• Jan 30th 2013, 11:08 PM
earboth
Re: Continuity problem
Quote:

Originally Posted by Paze
Find values of the constants a and b for which the following function is continuous but not differentiable.

$\displaystyle f(x)=\left\{\begin{array}{cc}ax+b,&\mbox{ if }x>0;\\sin2x, & \mbox{ if } x\leq 0\end{array}\right.$

In other words, the graph of the function should have a sharp corner at the pont (0,f(0)).

I'm not even sure what they are asking. This is part of a limit course and I am very familiar with limits.

1. Replace x by 0 in the appropriate part of the function. You'll get f(0) = 0

2. Therefore b must be zero.

3. The function has the slope f'(0) = 2

4. Thus for $\displaystyle a \ne 2$ the function is not differentiable at x = 0.

5. I've attached a sketch of the situation.
• Jan 30th 2013, 11:23 PM
hollywood
Re: Continuity problem
You will need to know at least something about derivatives to solve this problem (otherwise how can you be sure that it's not differentiable). The derivative of $\displaystyle \sin{2x}$ at 0 is 2, and the derivative of ax+b is a, so since they want it to be not differentiable, $\displaystyle a\ne{2}$.

For it to be continuous, the limit from the right (using ax+b) must be equal to the limit from the left (using $\displaystyle \sin{2x}$). You should be able to calculate those two limits and set them equal to get an equation for b.

- Hollywood

P.S. earboth beat me to the answer.
• Jan 30th 2013, 11:42 PM
Paze
Re: Continuity problem
I understand derivatives, I should probably have mentioned that.

However I don't understand why the derivative of sin2x at 0 is 2. The derivative of sinx is cosx so I assume the derivative of sin2x is cos2x...The derivative of sin(2*0) is 0 and the derivative of cos(2*0) is 1...How do you get 2?

EDIT: Something tells me I am supposed to be using chain/product rule. If so then I have to vent some frustration at these courses all over the net. They ask you to solve problems which you have not learned every single time when teaching derivatives.

EDIT2: Solved with chain rule. Jesus Christ, I like how these morons shoe-horn chain rule in the problems before properly teaching derivatives! Thanks sent to you both!
• Jan 31st 2013, 06:38 AM
hollywood
Re: Continuity problem
So it sounds like you understand why $\displaystyle a \ne 2$. Did you figure out the equation for b?

- Hollywood
• Jan 31st 2013, 10:23 AM
Paze
Re: Continuity problem
Quote:

Originally Posted by hollywood
So it sounds like you understand why $\displaystyle a \ne 2$. Did you figure out the equation for b?

- Hollywood

It seems to me that the function is continuous but not differentiable for any a except 2? In other words I'm getting y=(any a except 2)*x+0
Am I understanding this wrong?
• Feb 1st 2013, 10:54 PM
hollywood
Re: Continuity problem
That is correct - you need b=0 to make it continuous and $\displaystyle a\ne{2}$ to make it not differentiable.

- Hollywood
• Feb 1st 2013, 11:07 PM
Paze
Re: Continuity problem
Cool! Thanks for the help to both of you!