Question

The equation of a tangent, of slope -1 to the function of 1/(x-1). Really difficult. How to get k in y=-x+k.

Answer 1

I broke it down and got x^2-x-kx-k+1=0
Can't perform quadratic on this...

Answer 2

hi surji

Answer 3

\[Let y=\frac{ 1 }{x-1 },\frac{ dy }{dx }=\frac{ -1 }{\left( x-1 \right)^{2} }\]
also slope of tangent =-1
\[hence \frac{ -1 }{\left( x-1 \right)^{2} }=-1\]
\[\left( x-1 \right)^{2}=1,x-1=\pm 1 ,x=0,2\]
when x=0.\[y=\frac{ 1 }{ 0-1 }=-1\]
when x=2, \[y=\frac{ 1 }{2-1 }=1\]
therefore points are (0,-1) and (2,1)
slope is -1
i think now you can find the eq. of tangents.

Answer 4

my name is surjit

Answer 5

do you understood ? or should i solve

Answer 6

sorry still looking at it

Answer 7

dydx=−1(x−1)2 what is this?

Answer 8

ohh the derivative

Answer 9

yes it is derivative

Answer 10

why is it =+_1?
isnt 0 the only correct x?

Answer 11

\[(x−1)2=1,x−1=±1,x=0,2\]

Answer 12

\[why is x-1=1 or -1\]

Answer 13

Let g(x)=\( \dfrac{1}{1-x}\). Note that there is a vertical asymptote at x=1. There are also two places where \( g'(x)=-1\) ( see sketch below). Let f(x)= \( -x +k\). Since \( g'(x)=-1\) at 2 places, then it's possible for f(x) to also have a tangent of -1 at these two places and therefore have two different values for k.

The steps to solve this problem are as follows:

1) Find x where \( g'(x)=-1\) by taking derivative of g(x) and setting equal to -1
2) Substitute value of x obtained back into g(x) to determine where it has this slope.
There will be two places.
3) Set f(x)=g(x) for each x where slope is -1 to solve for the two constants (k and b)

Done.

I found that f'(x)=g'(x)=-1 at x = 2,y=1 and at x=0,y=-1.
$$
y_1(x)=-x+3\\
y_2(x)=-x-1
$$