help with extrema please

Question

hdrager · Answer

$$f(x)= \frac{ 2x-8 }{ x-1 }$$
$$f'(x)=\frac{ 6 }{ (x-1)^2 }$$
The you set it = 0
but idk what to do from there

sunnnystrong · Answer

Local or global?

hdrager · Answer

it doesn't mention any of that, it just says to find any relative extrema

sunnnystrong · Answer

alright well looks like you have solved for f prime.
now, when f'x = 0 --> you have a critical point. this is when the derivative is zero. aka you have a possible local max/min or global max/min.

After finding all of the critical points (when f'(x) is undefined or equal to zero) you need to do the first derivative text to classify all relative extrema.

hdrager · Answer

my problem is finding the critical point I think

hdrager · Answer

i'm not sure how to isolate x

sunnnystrong · Answer

Ps we will be looking only at the local extrema (relative extrema) as an interval hasn't been given. and alright well $$f'(x) = (6)/(x-1)^2 $$
Set this equal to zero

---> critical points are when f'(x) = 0 / f'(x) undefined so when is f'(x) undefined? when the denominator is equal to zero

sunnnystrong · Answer

Because the domain is all real numbers... we don't have any domain restrictions (important.) this derivative does not ever equal zero.. if you were to graph this you would see that as x-> infinity the limit is 0.

hdrager · Answer

okay thank you

hdrager · Answer

i got undefined but i didn't think it was correct

sunnnystrong · Answer

NP. & the function is only undefined when the denominator of the equation (x-1)^2 is equal to zero (it is a vertical asymptote). That is your only critical point. 

The first critical pt. you do a sign test and look at the signs of the values to the left and right of the critical pt. --> depending on how the signs change you either have a relative maximum or a relative minimum or neither

hdrager · Answer

okay thanks!