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&#039;(c) \Rightarrow f(b)-f(a)=(b-a)f&#039;(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&#039;(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:

<br /> \frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{2}<br />

(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!