Question

Determine if the function g(x)={x+1, x<1 and 2 sqrtx, x>1 is differentiable at x=1

Answer 1

Basically - do both parts of this piece-wise function have the SAME SLOPE at x=1?

Answer 2

ah okay let me do that know, could you check my answer to see if they are right?

Answer 3

Sure, go ahead!

Answer 4

okay cool thx!

Answer 5

I can do an example of a similar problem while you do that (if you want).

Answer 6

oh okay yea that be amazing thank you

Answer 7

Oh, wait, on thing about your function ... it should say should have one \(\le\) or \(\ge\) sign.

Answer 8

oh well it doesn't say so on my assignment

Answer 9

So your piece-wise function is

\(\displaystyle\color{black}{f(x) =\begin{cases}x+1 & x < 1\\2\displaystyle\sqrt{x} & x > 1\end{cases}}\)

Answer 10

yes

Answer 11

oh wait its g(x) but I feel that doesn't matter

Answer 12

it's weird that there is no point at x=1.

Answer 13

sooo what does that mean for answering this problem?

Answer 14

You want to know if the function has a slope at \(x=1\), however there is not even a point at \(x=1\).

Answer 15

Does that mean the same as differentiable at x=1?

Answer 16

In order for a function to be differentiable at \(x=c\) it must be continous at \(x=c\).

If \(f\) does not have a point at \(x=c\), then (all the more so) it is not differentiable at \(x=c\).

Answer 17

I also know there is a hole at (1, 2) based on the graph soo idk if that a factor in this.

Answer 18

so okay okay, but I have to solve it by g(1) or something like that

Answer 19

Well, if the piece wise function you are dealing with is:
\(\displaystyle\color{blue}{f(x) =\begin{cases}x+1 & x < 1\\2\displaystyle\sqrt{x} & x > 1\end{cases}}\)

And not
\(\displaystyle\color{black}{f(x) =\begin{cases}x+1 & x < 1\\2\displaystyle\sqrt{x} & x \color{red}{\ge} 1\end{cases}}\) or \(\displaystyle\color{black}{f(x) =\begin{cases}x+1 & x \color{red}{\le} 1\\2\displaystyle\sqrt{x} & x > 1\end{cases}}\)

Answer 20

then, you simply don't have a point at x=1.
(Your function is not continuous at x=1)

Answer 21

And if a function is NOT CONTINUOUS at a point, THEN it is NOT DIFFERENTIABLE at this point.

Answer 22

okay so how would I prove that while using the limit definition

Answer 23

You don't need to prove that your function is NOT DIFFERENTIABLE at x=1.

All you need is the fact that the function is NOT CONTINUOUS at x=1,
(that is - there is no point at x=1.)

Answer 24

«\(f\) is differentiable at a point x=c» is an equivalent of saying,
«\(f\) has a slope (a defined slope) at x=c»

Answer 25

If \(f\) does not even contain a point at x=c, then it certainly can not have a (defined) slope at x=c. (Right?)

Answer 26

and this is exactly what's going on with your function - it does not contain a point at x=1, and thus it cannot be differentiable at x=1.

Answer 27

okay sure it just that the material requires us to prove it using the definition of a limit but I will just write that down in my assignment.

Answer 28

Thank you for your help I have a few more if you dont mind helping me with it.

Answer 29

More support:

Definition: *A differentiable function of one real variable is a function whose derivative exists at each point in its domain.*

(and x=1, is apparently not even in the domain of \(f\))

Answer 30

Again, unless there was a typo and a function has a \(\le\) or \(\ge\) in one of the restrictions of the piece wise function (as I pointed out).

(because that would dictate continuity at x=1, since the parts are equivalent when x=1 .... if there was x=1 in the domain, which there isn't)

but, for other than that .... You Welcome!
~ Bye