Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. For every $x \geq 0$ prove that $\sqrt{x+1} - \sqrt{x} = \frac{1}{2\sqrt{x+\theta(x)}}$ for $\theta(x) \in (0,1)$. Also prove that $\theta(x) \in (\frac{1}{4}, \frac{1}{2})$ with $lim_{x \rightarrow 0} \theta(x) = \frac{1}{4}$ and $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}$

  2. Given a continuous real valued function $f$ on $[0,1]$, show that for some $a \in [0,1]$, $\int^1_0x.f(x)dx = \frac{1}{2}f(a)$.

In Q.1, the first part is the direct application of the Mean value theorem, what I don't get is how to limit $\theta(x)$ to $(\frac{1}{4}, \frac{1}{2})$. I have no clue how to approach Q.2.

Any clues or references are welcomed.

Thanks in advance.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

(1) $$ \sqrt{x+1}-\sqrt{x}=\frac{1}{2\sqrt{x+\theta(x)}}\implies \sqrt{x+\theta(x)}=\frac{1}{2(\sqrt{x+1}-\sqrt{x})}\implies\cdots $$ (2) $$ m\le f(x)\le M\implies\int^1_0 mx\,dx\le\int^1_0 x\,f(x)\,dx\le\int^1_0 Mx\,dx $$ and use the theorem of Bolzano.

share|improve this answer
2  
Thanks a lot for your help. –  Morty Apr 14 at 9:00

If we apply the mean value theorem for the function $f(x)=\sqrt{x}$ on the interval $[x,x+1]$ then we have $$\sqrt{x+1} - \sqrt{x} = \frac{1}{2\sqrt{a}}=\frac{1}{2\sqrt{x+(a-x)}}$$ where $a\in(x,x+1) $ and so $(a-x)=\theta(x)\in(0,1)$.

If you take the limit as x tends to zero and infinity then you get that $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}$ and $lim_{x \rightarrow \infty} \theta(x) = \frac{1}{2}.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.