Medium Hard continuity proof Tutorial Q6

[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.

 

[nuuskur]

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.

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 $x\in (-\delta,\delta)$.

 

[TanWu]

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.