Question

I have a function: (1)/(x-2)-3, i take the derivative and get : -(1)/(x-2)^2. Why are there no critical points?

Answer 1

Um....1/(x-2)^-3 is (x-2)^3....

Answer 2

its (1)/(x-2)-(3)

Answer 3

It's the same.

Answer 4

Thats the orginal function

Answer 5

there are critical points. what about when x=2? then you have a 0 on the bottom and it doesnt exist. your derivative is correct however.

Answer 6

well, my book says that there is only a vertical and horizontal asympotot at 2, and -3 respectively, but it says there are no point of inflecition or extrema

Answer 7

but isnt x=2 a critical point of the derivative

Answer 8

Am I missing something here, we appear to be talking about 2 completely different functions.

Answer 9

Critical point is 2. Critical point means only 'investigate further'; some thing could be going on here.

Answer 10

but in this particular function , there is nothing more i can deduce from the x=2 other than its a vertical asymptote correct?, perhaps thats why the book says that there are no critcal points

Answer 11

We don't know what the question is. Give us the full question.

Answer 12

exactly, there is a critical point at f(c) if f(c) does not exist or if f(c)=0. so there can be no point where the function is =0 because of your equation. it will get close to 0, but not get to 0. the only other places where it could be critical it is an asymptote instead

Answer 13

exxcellt, thank you

Answer 14

(1)/(x-2)-3 Does this mean 1 over (x-2)^-3 ?

Answer 15

no prob. sometimes it's helpful to graph it out too, in test situations where you cant get help.

Answer 16

no, it means (1/((x-2)^2))-3

Answer 17

without the ^2. my bad, thinking of the derivative

Answer 18

Ah, now I see...

Answer 19

The graph is not a simple function like a parabola, that gives max or min instantly. It just goes on forever.

Answer 20

right, its basically or is the 1/x graph shifted 2 units tot he right and then 3 units down

Answer 21

yup. you got it. so by seeing that you can figure out that there is no min and the point where it would be a max it doesnt exist. so that may help, hopefully. if it doesnt well then forget i said anything! lol