# For which a and b is the piececewise function continues everywhere

1. Nov 6, 2011

### Boik

The problem statement, all variables and given/known data:

Sorry this is a more accurate translation:
For which a and b does the function have a derivative everywhere

$$f(x) = \begin{cases} ax+b, & x≤1\\ x^2, & x>1 \end{cases}$$

I'd be grateful if someone explains to me the algorithm for solving these type of problems.

Last edited: Nov 6, 2011
2. Nov 6, 2011

### MaxManus

Both ax + b and x^2 are continues functions so all you have to do is set a and b so that ax+b = x^2 when x goes to 1.

3. Nov 6, 2011

### Deveno

first, we want f to be continuous (if f is not continuous, we'll have a hard time taking the derivative, unless it's a removeable discontinuity).

the problem obviously lies at x = 1, we should like the two definitions to agree there.

but that gives more than one possible choice for a and b, so we should also like the derivatives of ax+b and x2 to agree at 1.

so calculate d/dx(ax+b) and d/dx(x2) at x = 1, and take it from there...

(in each case, the reasoning behind making the two forms of f and f' agree at x = 1, is so that the left-hand limits and the right-hand limits both agree, ensuring the continuity of f in the first case, and the differentiablity of f in the second case).

4. Nov 6, 2011

### Boik

So basically the limits of both parts must be the same when x is approching the point and both parts must have the same derivative at that point in order for the function to have a derivative everywhere?

Hence lim(ax+b) as x->1, x<1 must be equal to lim(x^2) as x->1, x>1
and (ax+b)' = (x^2)'

I got that a = 2 and b = -1

Last edited: Nov 6, 2011
5. Nov 6, 2011

### Deveno

well, not exactly, but it's pretty clear f(x) = ax+b is differentiable anywhere to the left of x = 1, and f(x) = x2 is differentiable anywhere to the right of x = 1.

so the only place where the derivative might fail to exist, is at x = 1, which we can fix by choosing a and b wisely.