Medium Hard continuity proof Tutorial Q6

• TanWu
TanWu
Homework Statement
Throughout these Tutorial Q6 to Q9 let ##c: \mathbb{R} \rightarrow \mathbb{R}## be a differentiable function whose derivative ##c^{\prime}## is continuous at 0 with ##c^{\prime}(0)=1##
Relevant Equations
##c^{\prime}(0)=1##.
I am trying to solve (a) and (b) of this tutorial question.

(a) Attempt:

Since ##c'## is at ##c'(0) = 1##, then from the definition of continuity at a point:

Let ##\epsilon > 0##, then there exists ##d > 0## such that ##|x - 0| < d \implies |c'(x) - c'(0)| < \epsilon## which is equivalent to ##|x| < d \implies |c'(x) - c'(0)| < \epsilon## or ##x \in (-d, d) \implies 1 - \epsilon < c'(x) < 1 + \epsilon##

Which when comparing ##1 - \epsilon < c'(x) < 1 + \epsilon## to ##\frac{1}{2}<c^{\prime}(x)<\frac{3}{2}##, this means that ##\epsilon = \frac{1}{2}##, however, how is that possible i.e. why are we allowed to say ##\epsilon = \frac{1}{2}##? To me it seems like one is fudging the proof.

(b) Attempt:

##c## is a one-to-one function from the open interval ##(-d,d)## onto the open interval ##(-f(-d), f(d)## since it is differentiable which is equivlenet to saying that for every ##x \in dom(c)##, then there exists on and only one ##c(x) \in range(c)##

I express gratitude to those who help.

EDIT: Typos fixed.

Last edited:
General remark: please don't replicate the style '## d\in\mathbb R > 0 ##'.

By definition of continuity at ##0##, when we take ##\varepsilon =\frac{1}{2}##, we are guaranteed ## |c'(x)-1| <\frac{1}{2} ## for all ##x\in (-d,d)## for some (possibly very small) ##d>0##.

Your proof of injectivity is not correct.
TanWu said:
since it is differentiable which is equivlenet to saying that for every ##x \in dom(c)##, then there exists on and only one ##c(x) \in range(c)##
This is false. You are reproducing the definition of a "function" - for every ##x\in \mathrm{dom}(c)##, there exists exactly one element ##c(x)\in\mathrm{range}(c)##. It is not true that every function is differentiable.

You are given that the derivative is positive in ##(-d,d)##. What does that tell you about the behaviour of ##c## in the interval ##(-d,d)##?

edit: further remarks - to improve readability, it is helpful to distinguish between different kind of objects using different style of symbols. For example, we have a function ##c## for which there exists ## \delta >0## such that ##|c'(x)-1|< \frac{1}{2} ## for every ##x\in (-\delta,\delta)##.

Last edited:
TanWu and docnet
nuuskur said:
General remark: please don't replicate the style '## d\in\mathbb R > 0 ##'.

By definition of continuity at ##0##, when we take ##\varepsilon =\frac{1}{2}##, we are guaranteed ## |c'(x)-1| <\frac{1}{2} ## for all ##x\in (-d,d)## for some (possibly very small) ##d>0##.

Your proof of injectivity is not correct.

This is false. You are reproducing the definition of a "function" - for every ##x\in \mathrm{dom}(c)##, there exists exactly one element ##c(x)\in\mathrm{range}(c)##. It is not true that every function is differentiable.

You are given that the derivative is positive in ##(-d,d)##. What does that tell you about the behaviour of ##c## in the interval ##(-d,d)##?

edit: further remarks - to improve readability, it is helpful to distinguish between different kind of objects using different style of symbols. For example, we have a function ##c## for which there exists ## \delta >0## such that ##|c'(x)-1|< \frac{1}{2} ## for every ##x\in (-\delta,\delta)##.
Thank you Sir. You have solved by doubt about the proof. I think my other doubt is solved now: From (a), $$c'(x) > 0$$ for $$(-d,d)$$ this implies c is strictly increasing on $$(-d,d)$$ and thus, is one to one.

nuuskur

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