Krdytkyu

up vote 1 down vote favorite I'm going through a proof for the mean value theorem. We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$ . Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$ . The slope of said secant is: $$m=frac{f(b)-f(a)}{b-a}$$ That is clear. Now the proof I'm following defines $g(x)$ like so: $$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$ What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$ . calculus derivatives proof-explanation share | cite | improve this question asked 3 hours ago ...

up vote 10 down vote favorite 4 I am asking if whether or not a finite group acts freely (as homeomorphisms) on $mathbb R^n$. To answer in the negative, it suffices to show: for any homeomorphism $f$ such that $f^d=text{id}_{mathbb R^n}$, then $f$ has a fixed point. I am looking for a complete self-contained answer. algebraic-topology share | cite | improve this question edited May 13 '16 at 7:38 user99914 asked Jan 3 '13 at 0:21 Ash GX 536 3 18 ...