1. ## Continuity problem

Hi everyone, this is a problem from a previous exam. I manage to solve half of it, but am struggling with the second half.

f(x) = x2 + x - 1 for x ≥ 0
f(x) = a cos(x) + b sin(x) for x < 0

Find a and b so that f(x) is continuous and differentiable for all x = 0

Can someone help solve this?

Thank you all!

2. ## Re: Continuity problem

Originally Posted by Nora314
f(x) = x2 + x - 1 for x ≥ 0
f(x) = a cos(x) + b sin(x) for x < 0

Find a and b so that f(x) is continuous and differentiable for all x = 0
Well, I found the required a and b for some x = 0, but I am not sure about all of them.

Equate the values of the two functions at 0, as well as the values of their derivatives at 0. You'll get two equations in a and b.

3. ## Re: Continuity problem

I wrote it a bit weirdly, but the functions I wrote are actually supposed to be one function. Would what you said still work then? I found that a = 1, by investigating the limit as x approaches 0.

4. ## Re: Continuity problem

Originally Posted by Nora314
the functions I wrote are actually supposed to be one function. Would what you said still work then?
Yes.

Originally Posted by Nora314
I found that a = 1, by investigating the limit as x approaches 0.
x² + x - 1 = -1 when x = 0 and a cos(x) + b sin(x) = a when x = 0. Therefore a = -1. Now find the left and right derivatives at 0 and equate them.

5. ## Re: Continuity problem

Thanks for the help, I did that. Let me see, derivative on the right is 2x + 1 and on the left -sin x + b cosx. When I put them equal to each other and x = 0, I got b = 1.

In the answer key it says the opposite values, it says that a = 1 and b = -1. I manged to get that answer too by using another method. Perhaps it does not matter if the values of a and b are switched around ? (lol, I always think that).

Ok, I solved it both ways now do you know why the answers are switched around when we use this method?

And btw, I really like your signature ^^

6. ## Re: Continuity problem

Originally Posted by Nora314
Let me see, derivative on the right is 2x + 1 and on the left -sin x + b cosx.
Since a = -1, the left derivative is sin(x) + b cos(x), but since sin(0) = 0, the sign of sin(x) does not matter.

Originally Posted by Nora314
In the answer key it says the opposite values, it says that a = 1 and b = -1.
a = -1 and b = 1 gives the red graph, while a = 1 and b = -1 gives the blue one.

Perhaps the textbook authors thought that the left function is a sin(x) + b cos(x).

Originally Posted by Nora314
And btw, I really like your signature ^^
Thanks!

7. ## Re: Continuity problem

Thank you so so much for your help ^^ It makes a lot more sense now. It is really nice that you made a graph and everything!