lim f(x+h)-f(x) / h = -1 / (x-2)^2

h-> 0

 

-1 / (x-2)^2 = -1

 

I don't know what im doing help??

Use the quotient rule. Set the derivative equal to -1 to get the x value where the tanget is negative one. Then use y = mx+b to get the tangent line.

thats what i did

 

x= 0,2

 

i just dont get the y = mx+b part

Those are the x-coordinates of the points on the graph with slope -1. Then you find the y-coordinate using your equation [imath]y=\frac{1}{x-1}[/imath] so you get the complete coordinate. You can stick the x and y into y=mx+b, using -1 as the slope, and find out the equations (solving for b).

Ok now get the derivative at x = 2 and x = 0 then find the value of the function at x = 2 and x =0 then find b such that y = mx + b remembering that the derivative gives the slope, M the original function gives Y and you already found X. There will be two sepret tangent lines and two seperate b's

If the slope of the tangent and the slope of the derivative are the same thing then you are looking for when the derivative equals -1.

 

If I where you I would rewrite the equation as follows:

 

[math]\frac{1}{(x-1)}=(x-1)^{-1}[/math]

 

and then just solve it using the chain rule once you have the derivative remember a few things like:

 

[math]\frac{dy}{dx}=m[/math]

 

[math]y-y_{1}=m(x-x_{1})[/math]

 

So once you know when the derivative is equal to -1 you already have m and x. So all you need to know is what y is at the x points you find.

of course i forgot to include the -1

 

damn i make sure i actually dont get something before i ask it then i realize how dumb my mistakes are.. lol