Question

I am going over the chp in my book increasing and decreasing functions and the first derivative test. Could some clarify for me what it means when the book says that a function is differentiable on an certain interval( ie. {-2,8] or [pi,2pi]?

Answer 1

it means that the derivative exists for all numbers in that interval

Answer 2

as an example consider
\[f(x)=\sqrt[3]{x}\] whose derivative is
\[f'(x)=\frac{1}{3\sqrt[3]{x^2}}\]

Answer 3

It means that you cannot take the derivative of the function for values of x outside of the given interval.

So take \(\frac{1}{(x-2)(x-3)}\) This function is differentiable anywhere the denominator is not zero, so the intervals would be \[(-\infty,2)\cup (2,3) \cup (3,\infty)\] Where \(\cup \) means "union". :)

Answer 4

now this derivative certainly exists everywhere but at 0 so the original function is "not differentiable at 0" and therefore it would not be differentiable on (-2,8) for example

Answer 5

mathteacher1729, for your example the critical numbers would be 2 and 3, but also 0 right, cause the function is undefined there?

Answer 6

or can anyone answer that for me?

Answer 7

like, when i am using the first derivative test, i would also have to include any points where the derivative is undefined right

Answer 8

if a function is not defined for some number, its derivative certainly is not defined there

Answer 9

yes critical points are where the derivative is or undefined

Answer 10

HOWEVER

Answer 11

if the function is undefined there as well, you can safely ignore those points in finding the max or min. if the function has no value there, then of course it cannot have a max or min there

Answer 12

they "critical points" where the derivative is undefined but the function is not usually arise from something like the example i wrote above. where taking the derivative forces a denominator on you that was not in the original function

Answer 13

okay thanks

Answer 14

yw

Answer 15

Mathfever, the function I used as an example above is defined at zero just fine.

\[\frac{1}{(x-2)(x-3)}\]

If we let x = 0

\[\frac{1}{(0-2)(0-3)}=\frac{1}{(-2)(-3)}=\frac{1}{6}\]

Answer 16

right, but lets say we had a function that was undefined at zero, i woul dhave to also use it at an x-value along with the other critical numbers ti determine the test intervals right?