Question

help

Answer 1

What is your question? And yes I got your private message to me. Sorry I didn't respond right away.

Answer 2

This is a polynomial you mention.
Everything is continuous and smooth about it.

Answer 3

A continuous and smooth function is a differentiable one.

Answer 4

Do you know what I mean by smooth?

Answer 5

The slopes don't change suddenly.
It is a change that can be formulated.

Answer 6

Formulated by one function that is

Answer 7

Like y=|x| is not differentiable at x=0

Answer 8

the slopes change suddenly at x=0

Answer 9

we can formula the slopes on the left by coming up with the expression f'(x)=-1
and we can formulate the slopes on the right by coming up with the expression f'(x)=1

Answer 10

wouldn't you say the change from -1 to 1
is a very sudden change?

Answer 11

Do you think y=cos(x) is differentiable?

Answer 12

The changes are smooth and fact the slope formula for the curve that is y=cos(x) is y=-sin(x).
The function is also continuous (no holes no vertical asymptotes)

Both y=sin(x) and y=cos(x) are differentiable everywhere.


However y=tan(x) is not differentiable everywhere.
There are an infinite amount of places on the graph of y=tan(x) that aren't differentiable
y=sin(x)/cos(x)
we have vertical asymototes occuring at \[x=\pm \frac{\pi}{2} +2 n \pi \]
So the function isn't continuous at those x values
therefore not differentiable at those x values

Answer 13

I hope you have a better idea of what we consider a differentiable function.
If you don't please let me know.

Answer 14

Here is a tricky one
how about y=x^(2/3) ?

Answer 15

Also, g(x) = 3x^2 - x + 5
g'(x) = 6x - 1 which exists for ALL values of x. Therefore, g(x) is differentiable for all x.

Answer 16

Using what aum said
y=x^(2/3) is not differentiable at x=0
because
\[y'=\frac{2}{3 \sqrt[3]{x}}\]
notice y' is not not continuous at 0 mostly because you cannot divide by 0 :)

Answer 17

also if you graph y=x^(2/3) you should see a sharp turn at x=0

Answer 18

If this question has been answered you can close this one and open up a new post for the next question.

Answer 19

still need help

Answer 20

i don't know what you need help on unless you tell me.