Question

help: find the critical points: g(x) x+8/x-1

Answer 1

\[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]

Answer 2

\[\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }\]

Answer 3

\[\frac{ -9 }{ x ^{2} -2x+1 }\]

Answer 4

then I'm stuck here :/

Answer 5

Find x in: \[\frac{ -9 }{ x^2-2x+1 } = 0\]

Answer 6

Remember that the value of the derivative is the slope of the line of g(x) at that point so when the first derivative is equal to zero, the slope of the curve is zero and it is therefore a critical point.

Answer 7

yes so x=0?

Answer 8

that's what I did later..

Answer 9

No, when x = 0 then the function is equal to -9...find the value of x so that the dericative is equal to 0.

Answer 10

Also, since there is no solution for that equation (just realised :p) You can find at which point it is undefined.

Answer 11

yes, i end up with -9=0 can you show me?

Answer 12

it's undefined right? because the x is not there..

Answer 13

What do you mean?

Answer 14

unless it was something like -9x/ x^2-2x+1

Answer 15

then there would be an x.

Answer 16

but since -9 is alone... there is no critical point. right?

Answer 17

Correct, but there IS an asymptote in the graph so I'm not sure if you'd consider that a critical point in your class?

Answer 18

whats the asymptote?

Answer 19

At x=1

Answer 20

how did you find 1 again?

Answer 21

btw thank you

Answer 22

Set the bottom of the function equal to zero so x^2-2x+1 = 0 and solve for x.

No problem :D