Question

If g(x) = x^(2/3), show that g'(0) does not exist.

Answer 1

Have you calculated g'(x) yet?

Answer 2

@mathmate isn't that just

\[g'(x) = \frac{ 2 }{ 3*\sqrt[3]{x} }\]

Answer 3

because of the power rule

Answer 4

\(\Large \checkmark\)

Answer 5

@IrishBoy123 Oh and then it g'(0) would be undefined.

Answer 6

@WorldUnseen
exactly, well done! :)

Answer 7

Wow thanks everyone! @mathmate @IrishBoy123

Answer 8

sorta

the derivative is the limit

and here you come at it from different sides and you get different answers

so it's not that simple

Answer 9

@IrishBoy123 Yeah, I know, that's how it really should be done. I just don't know how to take the limit of that on both sides.

Answer 10

plotting it is always a good idea

Answer 11

@IrishBoy123 I know what the graph looks like, and it's pretty obvious why it can't exist, I just wanted to see how to explain it with limits/derivatives

Answer 12

well, its a 2 sided limit

different as you approach from either side

so the limit does not exist

Answer 13

@IrishBoy123 Could you explain to me how to get the exact values of those limits?

Answer 14

I know it should have to be as h approaches 0 on both sides (+ and -), but I don't really know how to simplify from there.

Answer 15

\(\Large \lim\limits_{x \to 0^-} \frac{ 2 }{ 3*\sqrt[3]{x} } = -\infty \)

\(\Large \lim\limits_{x \to 0^+} \frac{ 2 }{ 3*\sqrt[3]{x} } = \infty \)

Answer 16

@IrishBoy123 okay I see, thanks