Proving an Inequality: Using the Mean Value Theorem

[springo]

Homework Statement

I need to prove an inequality for 0 < x < +∞.

Homework Equations

The Attempt at a Solution

I guess it must be something with the mean value theorem, but I can't find what it is.

Thanks for your help.

Edit: Added range for x.

 

[gabbagabbahey]

This inequality is only true for a select range of x-values, so what range are you supposed to prove this for?

 

[springo]

Sorry forgot to say: 0 < x < +∞

 

[Mark44]

I don't see where restricting x to the positive reals does anything for you.

2 <= 3 + cos(x) <= 4, for all real x, so ln(3 + cos(x)) is defined for all real x, as well.
Taking logs, the inequality above becomes
ln(2) <= ln(3 + cos(x)) <= ln(4) = 2ln(2)

I don't see how the lower and upper bounds of the original inequality figure in, though.

 

[gabbagabbahey]

Hint: Let $f(x)=\ln (3+ \cos x)$...what does the mean value theorem have to say about $f(x)-f(0)$?

 

[springo]

With:

Right?But still, I can't think of anything but making that:

And then replacing, but still nothing.

 

[gabbagabbahey]

Perhaps we should go over the mean value theorem first...it says that for some $c \in [a,b]$
, the following will hold true:

$$\frac{f(b)-f(a)}{b-a}=f'(c) \Rightarrow f(b)-f(a)=(b-a)f'(c)$$

So looking at $f(x)-f(0)$ means $a=0$ and $b=x$ in the above relation right?...what does that give you?

 

[springo]

I thought $$b = 3+cos x$$ and $$a = 3+cos 0 = 4$$

So for: $$0 \leq l \leq x$$ we have:
$$ln(3+cos x) - ln(4) = \frac{x}{l}$$
Is that right so far?

 

[gabbagabbahey]

What happened to $f'(l)$?

 

[springo]

OK, I forgot that it's not $$ln(l)$$ but rather $$ln(3+\cos l)$$
So the derivative would be:
$$-\frac{\sin l}{3+\cos l}$$
So finally:
$$\ln(3+\cos x)-\ln 4 = \frac{-x\sin l}{3+\cos l}$$

 

[gabbagabbahey]

good, and what are the minimum and maximum values of $$\frac{-\sin l}{3+\cos l}$$ for any real $l$?

 

[springo]

The only way I can think of for doing that is:
$$\frac{1}{2}\geq\frac{1}{3+\cos l}\geq\frac{1}{4}$$

And:
$$1\geq-\sin l\geq-1$$

Therefore:
$$\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{4}$$

 

[gabbagabbahey]

Close, the actual relationship is:

$$\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{2}$$

(since -1/4 is actually greater than -1/2 not less than)

Given this, and the fact that you are restricted to positive x-values, what are the restrictions on $$\frac{-x\sin l}{3+\cos l}$$ ?

And hence what are the maximum and minimum values of $f(x)-f(0)$?

 

[springo]

OK, I think I got it now, but am I supposed to know that $$1\leq\ln4\leq\frac{3}{2}$$ in order to answer or is there any way around it?

 

[gabbagabbahey]

OK, I think I got it now, but am I supposed to know that $$1\leq\ln4\leq\frac{3}{2}$$ in order to answer or is there any way around it?

I think that's all there is to it.

 

[springo]

OK, thanks a lot!