Some simple Questios about Derivaties

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May 30th 2011, 02:08 AM
nadia321

Some simple Questios about Derivaties

Dear All,
I have a simple question which creates confusion for me

1. What is derivative of zero function? I know derivative of constant is zero and derivative of ZERO is zero. But the problem is derivative of ZERO Function at Zero becomes indeterminate form i.e
$f'(x)=\lim_{x \to \0} \frac{\mathrm{0-0} }{\mathrm{x-0} }$

2. We know that continuity is defined on some closed interval $[a, b]$ , but derivative is defined on open interval $(a, b)$ . Why?

3. In almost every book of calculus, theorem about test of increasing and decreasing function is given one way as:
If f is continuous on closed interval $[a, b]$ , and differentiable on $(a, b),$
(i) if $f'(x)>0,$ for all x in $(a, b)$ , then f is increasing
(ii) if $f'(x)<0$ , for all x in $(a, b)$ , then f is decreasing [/TEX].
To my knowledge converse of the theorem is also true i.e if f is increasing then $f'(x)>0,$ . In books why only way statement is written? or I am wrong.
May 30th 2011, 02:29 AM
Ted

1. Zero function is a constant also. No need for the limit definition.

2. Good question! It is impossible for the function to be differentiable on the end points because in order to be differentiable it should continuous first, and in order to be continuous it should be continuous from right and left. But the function is not defined at the right of b and the left of a. This means it is not continuous on b from right (or a from left), hence it is not differentiable.

3. Because this test is given as an application of the derivatives so it should be written so that the student will differentiate in order to test his understanding of differentiation.
and you are right. f is increasing (decreasing) if and only if f' > 0 (f' < 0).
May 30th 2011, 02:30 AM
Prove It

1. Use $\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ instead.
May 30th 2011, 10:40 AM
HallsofIvy

Haven't you noticed that every derivative is an indeterminant form?

$\lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$

Obviously, the denominator is going to 0. In order that the limit exist, it must be true that the numerator also goes to 0 (f is continuous).

In order to take the limit defining any derivative, you need to use this:
If f(x)= g(x) for all x except x= a, $\lim_{x\to a} f(x)= \lim_{x\to a}g(x)$

The derivative of the constant function, f(x)= 0 for all x, [tex]\frac{0- 0}{h}= 0[/itex] for all x except x= 0 so
$f'(x)= \lim_{h\to 0}\frac{0 - 0}{h}= \lim_{h\to 0} 0= 0$
May 30th 2011, 11:31 AM
Ackbeet

Quote:

Originally Posted by Ted

2. Good question! It is impossible for the function to be differentiable on the end points because in order to be differentiable it should continuous first, and in order to be continuous it should be continuous from right and left. But the function is not defined at the right of b and the left of a. This means it is not continuous on b from right (or a from left), hence it is not differentiable.

True for the two-sided derivative. You can actually define the derivative from the right at a left endpoint, and the derivative from the left at a right endpoint. In the limit definition of the derivative, you'd just use one-sided limits. I think some books may do it that way (or I could be mistaken). In that case, you could say that the function is differentiable on a closed interval if it's differentiable (two-sided) on the open interval, and differentiable from the right at the left endpoint, and differentiable from the left at the right endpoint.