1. The problem statement, all variables and given/known data I need to prove an inequality for 0 < x < +∞. 2. Relevant equations 3. 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.
This inequality is only true for a select range of x-values, so what range are you supposed to prove this for?
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.
Hint: Let [itex]f(x)=\ln (3+ \cos x)[/itex]...what does the mean value theorem have to say about [itex]f(x)-f(0)[/itex]?
With: Right? But still, I can't think of anything but making that: And then replacing, but still nothing.
Perhaps we should go over the mean value theorem first...it says that for some [itex]c \in [a,b] [/itex] , the following will hold true: [tex]\frac{f(b)-f(a)}{b-a}=f'(c) \Rightarrow f(b)-f(a)=(b-a)f'(c)[/tex] So looking at [itex]f(x)-f(0)[/itex] means [itex]a=0[/itex] and [itex]b=x[/itex] in the above relation right?....what does that give you?
I thought [tex]b = 3+cos x[/tex] and [tex]a = 3+cos 0 = 4[/tex] So for: [tex]0 \leq l \leq x[/tex] we have: [tex]ln(3+cos x) - ln(4) = \frac{x}{l}[/tex] Is that right so far?
OK, I forgot that it's not [tex]ln(l)[/tex] but rather [tex]ln(3+\cos l)[/tex] So the derivative would be: [tex]-\frac{\sin l}{3+\cos l}[/tex] So finally: [tex]\ln(3+\cos x)-\ln 4 = \frac{-x\sin l}{3+\cos l}[/tex]
good, and what are the minimum and maximum values of [tex]\frac{-\sin l}{3+\cos l}[/tex] for any real [itex]l[/itex]?
The only way I can think of for doing that is: [tex]\frac{1}{2}\geq\frac{1}{3+\cos l}\geq\frac{1}{4}[/tex] And: [tex]1\geq-\sin l\geq-1[/tex] Therefore: [tex]\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{4}[/tex]
Close, the actual relationship is: [tex] \frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{2} [/tex] (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 [tex]\frac{-x\sin l}{3+\cos l}[/tex] ? And hence what are the maximum and minimum values of [itex]f(x)-f(0)[/itex]?
OK, I think I got it now, but am I supposed to know that [tex]1\leq\ln4\leq\frac{3}{2}[/tex] in order to answer or is there any way around it?