Question

how do i find the derivative of
f(x)=5-IX-5I

Answer 1

well, the limit from the left and the right dont match when you get to x=5; so there is no 'derivative'

Answer 2

how would i find the critical numbers

Answer 3

but, ther eis a derivative for the left and a derivative for the right if anything...

the left = 1 and the right = -1

Answer 4

how?

Answer 5

recognize it as the absolute value function and notice the high point is at x=5

Answer 6

what is a u-substitiution?

Answer 7

it is where you rename part of an expression as u='whatever you want to substtute u with'

Answer 8

its a way to clean up a messy expression

Answer 9

how would it work here? with this problem, if you dont mind explaining

Answer 10

cause i see that in regards to this function it can be writen as : \[-(x-5)/(IX-5I)\]

Answer 11

u = x-5 is the most likely substitution

5 - |u| is your expression....

your still left with trying to determine the derivative of an underivable equation that would have to be split up into seperate domains

Answer 12

\[f(x)=\left\{\frac{5-u: u>5}{5+u:u<5} \right\}\]

Answer 13

what happens if the original function is undefined at a value

Answer 14

will the derivative be undefined at that same value

Answer 15

as long as the limit from the left and right converge to a single number; then its good ... if i reall correctly

Answer 16

what happens with the function : 1+1/x

Answer 17

I mean x+1/x

Answer 18

remember the definition of a derivative is:

\[\lim_{h ->0}\cfrac{f(x+h)-f(x)}{h}\]
if the limit doesnt exist, the derivative dont exist

Answer 19

just by looking at the original function, i can automatically say that there is a vertical asympote at 0 correct.

Answer 20

yes, thats a fair assumption

Answer 21

now in this case x=0, would not be considered a critical number right, since both the original function as well as the derivative is undefined here. There would be a discontinuity at 0.

Answer 22

But, lets say that the function was still defined at a number but the derivative is not, then that number would still be a critical number right?

Answer 23

yes.