Proving differentiability for inverse function on given interval

• TanWu
TanWu
Homework Statement
Prove that ##e:(c(-d), c(d)) \rightarrow(-d, d)## is differentiable by computing ##e^{\prime}(s)## from the definition for each ##s \in (c(-d), c(d))##. You may use the fact ##\frac{2}{3} < \frac{e(t) - e(s)}{t - s} < 2## for ##t \neq s \in (c(-d), c(d))## without proof.
Relevant Equations
##\lim_{t \to s} \frac{2}{3} < \lim_{t \to s} \frac{e(t) - e(s)}{t - s} < \lim_{t \to s} 2##
I am trying to solve (a) and (b) of this question.

(a) Attempt

We know that ##\frac{2}{3} < \frac{e(t) - e(s)}{t - s} < 2## for ##t \neq s \in (c(-d), c(d))##

Thus, taking the limits of both sides, then

##\lim_{t \to s} \frac{2}{3} < \lim_{t \to s} \frac{e(t) - e(s)}{t - s} < \lim_{t \to s} 2##

Or equivalently,

##\frac{2}{3} < \lim_{t \to s} \frac{e(t) - e(s)}{t - s} < 2##

##\frac{2}{3} < e'(s) < 2##

Thus, using the squeeze principle, then ##e'(s)## is bounded between ##\frac{2}{3}## and ##2##, then the derivative exists for ##t \in (c(-d), c(d))## and thus we have proved that ##e(s)## is differentiable on the required interval. Are we allowed to do that?

(b) Attempt

Since ##c(x)## is ##C^1## on (-d, d) and as we proved in (a), then e must be ##C^1## on ##(e(-d), e(d))## as it is differentiable and thus continuous. This seems somewhat too trivial of a question to me.

I express gratitude to those who help.

nuuskur
The squeeze principle requires the smallest and largest things to become equal in the limit. 2/3 and 2 are not equal, no matter how close t and s are...

TanWu
Office_Shredder said:
The squeeze principle requires the smallest and largest things to become equal in the limit. 2/3 and 2 are not equal, no matter how close t and s are...
Thank you Sir. I apologize over my mistake. Is my proof correct thought without reference to the squeeze principle thought? Like using the fact the derivative is bounded? I cannot think of any other ways of proving it.

I don't really understand what the point of letting you use that inequality is (especially since you proved it in question 7, so it's not like they're doing you a great favor). I honestly do not know what they expect the answer to look like. I assume you're supposed to take that ratio and express it in terms of c and then compute something that looks similar to the derivative of c.

TanWu

• Calculus and Beyond Homework Help
Replies
12
Views
346
• Calculus and Beyond Homework Help
Replies
2
Views
364
• Calculus and Beyond Homework Help
Replies
2
Views
352
• Calculus and Beyond Homework Help
Replies
5
Views
174
• Calculus and Beyond Homework Help
Replies
1
Views
772
• Calculus and Beyond Homework Help
Replies
2
Views
370
• Calculus and Beyond Homework Help
Replies
2
Views
661
• Calculus and Beyond Homework Help
Replies
2
Views
281
• Calculus and Beyond Homework Help
Replies
7
Views
471
• Calculus and Beyond Homework Help
Replies
6
Views
393