Finding tangent of F(x) if we know tangent passes through a point f(x) does not...

**Saibot** · February 19th 2009, 05:38 PM

I cant see to understand how to do this question.

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

Find tangent if tangent goes through (1,1)

**skeeter** · February 19th 2009, 07:04 PM

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

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

point not on curve $(1,1)$

slope between points ...

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

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

$\frac{2 - (x-1)^2}{(x-1)^3} = -\frac{4}{(x-1)^3}<br />$

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

$(x-1)^2 = 6$

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

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

note that two tangent lines exist with the desired properties.

finish up by finding the tangent line equations

**Saibot** · February 19th 2009, 07:22 PM

Actually i did it this way and i got something totally different. What am i doing wrong?

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

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

$y=mx+b$

$b= 1-m$

$y=f(x)<br /> f'(x)=m$

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

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

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

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

What am i doing wrong?

i was following these instructions...:
http://www.karlscalculus.org/l4_5.html

**skeeter** · February 20th 2009, 05:28 AM

Originally Posted by Saibot

Actually i did it this way and i got something totally different. What am i doing wrong?

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

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

$y=mx+b$

$b= 1-m$

$y=f(x)<br /> f'(x)=m$

$\frac{2}{(1-x)^2}=(\frac{-4}{(1-x)^3})x + 1 + \frac{4}{(1-x)^3}$ this step is fine.

$\frac{2(1-x)-4x+x-4}{(1-x)^3}=0$ this step has errors. should start out as ...

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

i was following these instructions...:
Karl's Calculus Tutor - Alternative Solution to Tangent Problem

first, I'll admit to a mistake ... I "saw" the original function as $f(x) = \frac{2}{(x-1)^2}$ instead of $f(x) = \frac{2}{(1-x)^2}$
(pardon my dyslexia)

however ...

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

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

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