Differentiable functions proof

  • #1
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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!!
 

Answers and Replies

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

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