I cant see to understand how to do this question.
f(x) = 2/(1-x)^2
Find tangent if tangent goes through (1,1)
$\displaystyle f(x) = \frac{2}{(x-1)^2}$
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
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
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.