Reasons why for the study of the derivability of a piecewise

[dunkelheit]

Is there a particular reason why, for a piecewise defined function, for the study of the continuity/derivability we must use the definition in the different points where the function is defined by pieces and we cannot use, in the case of derivability, the derivation rules instead of the limit of the difference quotient?
I don't see from the theory why we must do this.
Thanks.

 

[Deleted member 4993]

Is there a particular reason why, for a piecewise defined function, for the study of the continuity/derivability we must use the definition in the different points where the function is defined by pieces and we cannot use, in the case of derivability, the derivation rules instead of the limit of the difference quotient?
I don't see from the theory why we must do this.
Thanks.

You say:

why we must do this. ​

Please provide an example (mathematical - no in words) of "doing this".

 

[dunkelheit]

Thanks for your answer. I mean that if I have to study the derivability of

[MATH]\begin{cases} f(x)=x^2, \text{if} \ x \leq 0 \\ \sqrt{1-x}, \ \text{if} \ 0<x<1 \\ \ln x, \ \text{if} \ x>0 \end{cases}[/MATH]​

I must use the definition of derivability (the limit of the difference quotient) in the points [MATH]x=0[/MATH] and [MATH]x=1[/MATH] instead of taking the derivative in this way

[MATH]\begin{cases} f(x)=2x, \text{if} \ x \leq 0 \\ -\frac{1}{2\sqrt{1-x}}, \ \text{if} \ 0<x<1 \\ \frac{1}{x}, \ \text{if} \ x>0 \end{cases}[/MATH]​

I would like to know why we must use the definition in the union points [MATH]x=0[/MATH] and [MATH]x=1[/MATH] of the curves.

 

[Steven G]

I suspect that you have a typo and it should say that f(x) = ln(x) if x>1 and 1/x if x>1

First of all you are given that f(x) = x^2 if x≤0, then you say f(x) = 2x if x≤0. I doubt that statement is true.

Seriously, if someone tell you that you must use the definition of derivative rather than the short cut method you should be asking them this question. You do NOT have to use the derivative definition to find f'(x) for any point.

Now to determine if f'(x) exists at x=0 and x=1 you first have to determine if the function is continuous at these points then you can look into the derivatives. If f(x) is not continuous at one or both points then there is no derivative at that (those) point(s) of discontinuity.

Now if the function is in fact continuous at x=0 and/or x=1 then you must make sure that the derivative from both sides of x=0 and x=1 give the same slope.

For example ASSUMING that f(x) is continuous at x=0, then the left hand deviate will be 2*0=0 and the right hand derivative would be -1/2. Since the slopes are different we will have a corner point at x=0 (if f(x) was continuous at x=0)

In your case f(x) is discontinuous at both x=0 and x=1.

 

[JeffM]

I probably do not understand your question because I am in my very own dunkelheit about what you mean by "derivability," which is a perfectly good word in English but differs in meaning from the mathematical meanings in English of "derivative," "differentiable," and "differentiability." (Here "bility" is a suffix similar to the German "keit" that transforms an adjective into a noun.)

But today has been weird enough that I shall chance an answer (chance because standard analysis is not at all my thing; to my taste, Weierstrass was a mathematician the way Ludendorff was a general: push forward no matter how ugly it gets.)

At points other than the division points, the general formulas for derivatives apply because those formulas were derived on the assumption of an open interval. Thus, being rigorous, they cannot apply to an endpoint of a closed interval. Thus, we must revert to the basic definition of the limit of a Newton quotient. If I am wrong, I am confident that pka, Halls, or my friend Subhotosh will soon correct me.

Let me apologize again in advance. I may not understand your question, and I have done my best to erase from memory whatever analysis I was forced to learn. My answer may be complete garbage.

PS There are two important diferences between W and L: the former was successful and murdered only beauty.

 

[Steven G]

Assuming continuity at a point x=a, then the derivative exits at x=a if the left hand limit equals the right hand limit at x=a. But you can use the whatever short cut rules to find the left hand derivative and the right hand derivative.

For example, suppose for x< 7, f(x) = x^3 and for x>7, f(x) = 48x+7.

By design f(x) is continuous at x=7. The left derivative at x=7 is 3x^2 evaluated at 7 which is 147. The right derivative at x=7 is 48. Since 48 and 147 are not the same then the derivative at x=7 does not exist. There is a corner point there.

I am fairly sure that what I said is true. However there may be a counterexample to my claim. I always have doubts after reading Counterexamples in Analysis by Gelbaum. It is a must read but it was hard on me seeing all those counter examples!