I cant see to understand how to do this question.

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

Find tangent if tangent goes through (1,1)

- Feb 19th 2009, 05:38 PMSaibotFinding tangent of F(x) if we know tangent passes through a point f(x) does not...
I cant see to understand how to do this question.

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

Find tangent if tangent goes through (1,1) - Feb 19th 2009, 07:04 PMskeeter
$\displaystyle f(x) = \frac{2}{(x-1)^2}$

point on curve ... $\displaystyle \left(x , \frac{2}{(x-1)^2}\right)$

point not on curve $\displaystyle (1,1)$

slope between points ...

$\displaystyle \frac{\frac{2}{(x-1)^2} - 1}{x-1} = \frac{2}{(x-1)^3} - \frac{1}{x-1} = \frac{2 - (x-1)^2}{(x-1)^3}$

$\displaystyle f'(x) = -\frac{4}{(x-1)^3}$

$\displaystyle \frac{2 - (x-1)^2}{(x-1)^3} = -\frac{4}{(x-1)^3}

$

$\displaystyle 2 - (x-1)^2 = -4$

$\displaystyle (x-1)^2 = 6$

$\displaystyle x = 1 \pm \sqrt{6}$

$\displaystyle y = \frac{1}{3}$ for both x-values

note that two tangent lines exist with the desired properties.

finish up by finding the tangent line equations - Feb 19th 2009, 07:22 PMSaibot
Actually i did it this way and i got something totally different. What am i doing wrong?

$\displaystyle f(x)=\frac{2}{(1-x)^2}$

$\displaystyle f'(x)=\frac{-4}{(1-x)^3}$

$\displaystyle y=mx+b $

$\displaystyle b= 1-m $

$\displaystyle y=f(x)

f'(x)=m$

$\displaystyle \frac{2}{(1-x)^2}=(\frac{-4}{(1-x)^3})x + 1 + \frac{4}{(1-x)^3}$

$\displaystyle \frac{2(1-x)-4x+x-4}{(1-x)^3}=0$

$\displaystyle \frac{-2-5x}{(1-x)^3} = 0$

$\displaystyle x= \frac{-2}{5}$

What am i doing wrong?

i was following these instructions...:

http://www.karlscalculus.org/l4_5.html - Feb 20th 2009, 05:28 AMskeeter
first, I'll admit to a mistake ... I "saw" the original function as $\displaystyle f(x) = \frac{2}{(x-1)^2}$ instead of $\displaystyle f(x) = \frac{2}{(1-x)^2}$

(pardon my dyslexia)

however ...

$\displaystyle \frac{2}{(x-1)^2} = \frac{2}{(1-x)^2}$

so, my solution should be the same. I'll have to work on it and try to get back to you.