- #1

TanWu

- 17

- 5

- Homework Statement
- Throughout this 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##

The tutorial question I am working on is,

(a) Attempt

We can use mean value theorem since

##(c: \mathbb{R} \rightarrow \mathbb{R}~countinity ) \implies (c: [-d, d] \rightarrow [c(-d), c(d)]~countinity)##

Thus ##c: [-d, d] \rightarrow [c(-d), c(d)] ## is differentiable on ##(-d, d)##, then there exists ##b \in (-d, d)## such that ##c'(b) = \frac{c(-d) - c(d)}{-d - d} = \frac{f(-d) - f(d)}{-2d}##

Using result from Q6,

##b \in (-d, d) \implies \frac{1}{2} < \frac{f(-d) - f(d)}{-2d} < \frac{3}{2}##

Not sure how to prove from here.

(b) Attempt

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

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

However, this is far from the expression that I am trying to prove.

(c) Nothing yet (may rely on (a) and (b))

I express gratitude to those who help

(a) Attempt

We can use mean value theorem since

##(c: \mathbb{R} \rightarrow \mathbb{R}~countinity ) \implies (c: [-d, d] \rightarrow [c(-d), c(d)]~countinity)##

Thus ##c: [-d, d] \rightarrow [c(-d), c(d)] ## is differentiable on ##(-d, d)##, then there exists ##b \in (-d, d)## such that ##c'(b) = \frac{c(-d) - c(d)}{-d - d} = \frac{f(-d) - f(d)}{-2d}##

Using result from Q6,

##b \in (-d, d) \implies \frac{1}{2} < \frac{f(-d) - f(d)}{-2d} < \frac{3}{2}##

Not sure how to prove from here.

(b) Attempt

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

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

However, this is far from the expression that I am trying to prove.

(c) Nothing yet (may rely on (a) and (b))

I express gratitude to those who help

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