Differentiable functions proof

[dancergirlie]

Homework Statement

Consider a function g : (a, b)-->R. Assume that g is differentiable at some point c in
(a,b) and that g'(c) is not = 0. Show that there is a delta > 0 so that g(x) is unequal to g(c) for all x in V_delta(c)\{c}intersect(a,b)

Homework Equations

The Attempt at a Solution

Alright, this is what I tried so far:

Since g is differentiable at c that means that

g'(c)=lim_x-->c (g(x)-g(c)/x-c)

since we are assuming that g'(c) is unequal to 0 that means that
lim_x-->c (g(x)-g(c)/x-c) is unequal to zero and thus g(x)-g(c) is unequal to zero
and therefore g(x) is unequal to g(c).

I don't really know where the delta comes in though, am I supposed to use the definition of a limit to show that there exists an epsilon greater than zero so that for |g'(c)-L|<epsilon?

Any help would be great!

 

[HallsofIvy]

Homework Statement

Consider a function g : (a, b)-->R. Assume that g is differentiable at some point c in
(a,b) and that g'(c) is not = 0. Show that there is a delta > 0 so that g(x) is unequal to g(c) for all x in V_delta(c)\{c}intersect(a,b)

Homework Equations

The Attempt at a Solution

Alright, this is what I tried so far:

Since g is differentiable at c that means that

g'(c)=lim_x-->c (g(x)-g(c)/x-c)

since we are assuming that g'(c) is unequal to 0 that means that
lim_x-->c (g(x)-g(c)/x-c) is unequal to zero and thus g(x)-g(c) is unequal to zero
and therefore g(x) is unequal to g(c).

I don't really know where the delta comes in though, am I supposed to use the definition of a limit to show that there exists an epsilon greater than zero so that for |g'(c)-L|<epsilon?

Any help would be great!

The fact that $\lim_{x\to c} (g(x)-g(c))/(x-c)= g'(c)$ is not 0 does NOT imply that (g(x)- g(c))/9x-c) is not 0! It does mean that you can take $\epsilon$ to be |g'(c)/2|[/itex] and then, for some $\delta> 0$ if $|x-c|< \delta$, $||g(x)- g(c))/(x-c)- g'(c)|< |g'(c)|/2$. Write that as $-|g'(c)|/2< (g(x)- g(c))/(x-c)- g'(c)< |g'(c)|/2$ and see where that leads.