differentiability study at a point

[toniuck]

Hello guys, I have the following question:

Say we have a function f:R->R, f(x)=y, if x is from [a, +infiniti) and f(x)=z, if x is from (-infinity, a), where y and z are two differentiable functions on [a, =infinity) and respectively ( -infinity, a).
The question is to study if f is differentiable in a.

I know I can solve this by calculating :
(f(x)-f(a))/(x-a) when x approaches x from left and from right and say that f is differentiable at a if those two limits are equal.

However, I was wondering if a different approach would work just as find. We can derivate f on those two intervals (without a) and get
f'(x)=y' , if x is from (a, +inf)
f'(x)=z', if x is from (-inf, a).
Now, the question is:
If we calculate limit when x approaches a from right from y' and then limit when x approaches a from the left from z' and find that those two limits are equal, can we say that f is differentiable at a? Or do we need to use the first proof?
(f is not necessarily f:R->R but i wanted to take a general case)

I will post an example to make my question more clear:
f:[0, infinity)->R , f(x)=(x-1)/(x^(1/2)-1), if x is from [0, 1) and
f(x)=x^2-x+2, if x is from (1, infinity)

Question: Study if f is differentiable at 1.

So, can we solve it as it follows?

f'(x)=1/(2*x^(1/2)), if x is from (0, 1)
f'(x)=2x-1, if x is from [1, inf)
then solve limit when x approaches 1 from the left from 1/(2*x^(1/2)) and then limit when x approaches 1 from the right from 2x-1 and see if those two are equal.
In this case, they are not, as the first is 1/2 and the other one is 1, so f is not differentiable at 1.
Is this proof correct? The answer is correct, because is the same answer we get by using the first proof. If the proof is not correct, could you please give me an example of a function where we get different results from these two methods?

Thanks a lot, and sorry for my presentation errors as English is not my native language and I do not study math in English either.

 

[Steven G]

A lim as x-> c of g(x) exist iff the left hand limit = the right hand limit.
Now the derivative by definition is a limit so the left hand derivative must equal the right hand limit.

So basically you need to verify if 'both' derivative equal one another when x=1.

 

[Ishuda]

First, let's be a little bit formal. For example, you have a problem with your example since, as stated, f(x) is not defined at 1. But, since it has a removable singularity, we can remove the singularity, by defining
g(x) = x^1/2 + 1; 0\(\displaystyle \le x \lt\) 1
It removes that problem since g (and all of its derivatives) and f (and all of its derivatives) are equal on the interval (0,1).

In the same sense, a derivative may have a removable discontinuity if it is defined in two disconnected intervals such as, for example, (a,b) and (b,c). Only one sided derivatives can be defined because the function is only defined on the one side. However, as you mentioned, if f' exists on (a,b) and on (b,c) and if
\(\displaystyle \lim_{x \to b-} f'(x) = lim_{x \to b+} f'(x) = f'_a\)
we may remove that discontinuity by defining \(\displaystyle f'(a) = f'_a\). Note that one definition of a derivative
\(\displaystyle f'(a) = lim_{h \to 0} \frac{f(a+\frac{h}{2}) - f(a-\frac{h}{2})}{h}\)
may not work in this case.

As an example, let's doctor up yours
f(x) = \(\displaystyle 2\, x^\frac{1}{2}\, +\, 1;\, 0\, \lt\, x\, \lt\, 1 \\x^2\, -\, x + 3;\, 1\, \lt\, x \lt\, \infty \)
f(1) = 3
f'(1) = 1
is continuous as well as having a continuous derivative.