How do you get from -(2)/((1+x)^2)=-1 to (1+x)^2=2? Does multiplying both side by negative 2 satisfie this? Wouldn't the left side become (4)/((1+x)^2)? If you want the entire problem, I can post it, but basically, I had to derive a function, f, and got -(2)/((1+x)^2), and now I have to find the value, C guarenteed by the mean value theorm, so I set f'(x) = to the slope of the secant line. Now I'm trying to solve for it.

Question

anonymous · Answer

How do I get $$-\frac{ 2 }{(1+x)^{2} }=-1$$
to be
$$(1+x)^{2}=2$$
? Multiply both sides by -2? Wouldn't that give me
$$\frac{ 4 }{ (1+x)^{2} }=2$$

anonymous · Answer

The question is:
Given the function $$f(x)=\frac{ 1-x }{ 1+x }$$
what is the slope of the secant line through the points (0,f(0)) and (1,f(1))? Find the value c between 0 and 1 guaranteed by the mean value theorm.

I got:
f(0)=1 and f(1)=0, so Msec=-1. $$f'(x)=-\frac{ 2 }{ (1+x)^{2} }$$
So 
$$-\frac{ 2 }{ (1+x)^{2} }=-1$$
What to do from here?