3
$\begingroup$

Using the mean value theorem prove the below inequality.

$$\frac{1}{2\sqrt{x}} (x-1)<\sqrt{x}-1<\frac{1}{2}(x-1)$$ for $x > 1$.

I don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.

I really wish someone could give a very simple solution.

$\endgroup$
6
$\begingroup$

Apply the mean value theorem to the function $f(t) = \sqrt{t}$ on the interval $[1,x]$ to deduce

$$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$$

for some $c \in (1,x)$. Use the fact that

$$\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$$

to conclude.

$\endgroup$
  • $\begingroup$ how did you know $$f(t)=\sqrt{t}$$? can you show me a different approach too? if possible. $\endgroup$ – Sherlock Homies May 30 '15 at 21:27
  • 3
    $\begingroup$ @SherlockHomies note $\sqrt{x} - 1 = \sqrt{x} - \sqrt{1} = f(x) - f(1)$, where $f(t) = \sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\frac{1}{2\sqrt{x}} < \frac{1}{\sqrt{x} + 1} < \frac{1}{2}.$$ $\endgroup$ – kobe May 30 '15 at 21:30
  • $\begingroup$ wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? $\endgroup$ – Sherlock Homies May 30 '15 at 21:49
  • 1
    $\begingroup$ @SherlockHomies multiply the inequalities $\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$ by $x - 1$ and use the fact $\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$ to get the desired inequalities. $\endgroup$ – kobe May 30 '15 at 21:52
1
$\begingroup$

By the MVT, there is $c \in ]1,x[$ such that $$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x-1), \qquad \qquad \left[f(b)-f(a) = f'(c)(b-a)\right]$$ so you use that: $$c < x \implies \sqrt{c} < \sqrt{x} \implies 2 \sqrt{c} < 2\sqrt{x} \implies \frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{c}}$$ to get one side, and use that: $$c > 1 \implies \sqrt{c} > 1 \implies 2\sqrt{c} > 2 \implies \frac{1}{2\sqrt{c}}< \frac{1}{2}$$ to get the other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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