Question

Can someone plz confirm a quick calculation with me plz??

Answer 1

So i have a function of \[F’(x) = \frac{ 1 }{ 3(x ^{\frac{ 2 }{ 3 }} )}\] so any value of x less than one will result in an undefined function true?? In other words, the function is defined for any value of x, except x>0.

Answer 2

I’m having a hard time plugging that into my old calculator :)

Answer 3

Sorry i meant less than 0...and not one..

Answer 4

Cube root values can be negative

So the exception is just 0

Answer 5

Actually nvm... i think i got it.. if i write that as cube root .....the negative values will become positive because of that square power....

Answer 6

But when i plugged in negative values into x^2/3......i was getting undefined..maybe because i need to put them into parentheses.....

Answer 7

X can’t be zero... because that’d make the entire function undefined... so there’s a vertical asymptote at x = 0... i’m Plugging in x values from both spectrums of x = 0 into the first derivative to see the behaviour of the original function.. trying to figure out where it’s increasing and where it’s decreasing. :)