Let f be a real continuous function. Which of the following statements are true and which are false? Give proofs or counter examples to justify your answers.

1. If f is a bijection and is also differentiable then its inverse is differentiable.

I don't know how to start this part of the question, any ideas?

3. If f is differentiable with f'(x) not equal to zero for any x, then f is a bijection.

I think this is true, I have used Rolles theorem to prove it must be injective but cannot prove it is surjective, I think I will need to use IVT but am not sure how.

4.If f is differentiable at c then $\lim_{x \to 0} \frac{f(c+h) -f(c-h)}{2h}$ exists and is equal to f '(c).

I think this is true but don't know how to prove it, any ideas?

2. 3.) What about $f(x)= e^x$? This is definitely injective but it's not surjective onto the real numbers because $f$ maps $\mathbb{R}$ onto $(0,\infty)$, not $\mathbb{R}$ onto $\mathbb{R}$

3. 4.) Use L'Hopital's Rule:

$\lim_{h\to 0}\frac{f(c+h)-f(c-h)}{2h} = \lim_{h\to 0}\frac{f'(c+h)+f'(c-h)}{2} = \frac{f'(c)+f'(c)}{2} = f'(c)$

4. What does cannot be ordered mean?

5. The complex numbers are not an ordered field.

Proof. Assume it is an ordered field. Then either $i>0$ or $i<0$ (because we know $i\neq0$). If $i>0$ then $i^2>0^2 \implies -1>0$, which is clearly false, so $i\not>0$. If $i<0$, then $-i>0$ and $(-i)^2>0^2 \implies -1>0$, which is clearly false, so $i\not<0$. Thus, none of the following are true: $i>0, i<0$, or $i=0$. Thus $\mathbb{C}$ is not an ordered field.

6. Does anybody know how to do 1?

7. Originally Posted by C.E
What does cannot be ordered mean?
The statement you're asking about is part of a signature and has nothing to do with your questions.

8. Originally Posted by C.E
Does anybody know how to do 1?
Originally Posted by C.E

Let f be a real continuous function. Which of the following statements are true and which are false? Give proofs or counter examples to justify your answers.

1. If f is a bijection and is also differentiable then its inverse is differentiable.

If the derivative of a function has a zero at any point, then the derivative of the inverse function cannot be defined at that point. This can be restated as the inverse function will not be differentiable at that point. Intuitively, the slope of the inverse function will be vertical and undefined at that point. Algebricaly, the inverse derivative can be computed with the Inverse function theorem, where the derivative of the original function is in the denominator of the formula. This theorem can be derived with the chain rule, and it is probably a topic that you are moving toward in your studies.

The inverse of a differentiable bijection is not necessarily differentiable, e.g. the inverse of $f(x) = x^3, f^{-1}(x) = x^{1/3}$ is not a differentiable because the derivative of f(x) equals 0 at x = 0 (and hence its inverse is not differentiable at x = 0). In more advanced terminology, this is an example of a homeomorphism that is not a diffeomorphism. (Counter example here derived from diffeomorphisms.)

Inverses and derivatives topic in invertible function.

9. Originally Posted by redsoxfan325
The complex numbers are not an ordered field.

Proof. Assume it is an ordered field. Then either $i>0$ or $i<0$ (because we know $i\neq0$). If $i>0$ then $i^2>0^2 \implies -1>0$, which is clearly false, so $i\not>0$. If $i<0$, then $-i>0$ and $(-i)^2>0^2 \implies -1>0$, which is clearly false, so $i\not<0$. Thus, none of the following are true: $i>0, i<0$, or $i=0$. Thus $\mathbb{C}$ is not an ordered field.

uhm.. i dont get it, $(-i)^2>0^2 \implies -1<0$

it goes just like $(2)>0$ so $(-2)<0$ but $(-2)^2<0^2 \implies 4<0$

clearly false
you can't sqaure negative numbers in this occasion. But, i is not a number. Confused now :P Hard to remember stuff i learnt last summer when since then i learnt almost everything maths need till these $\int_{a}^{b}$ (i don't know it's name in english)