Ahh... Good ol' math (real analysis) help from all you nice folks...

[dvdrdiscs]

Use the mean value theorem to prove that (x-1)/x < lnx < x-1 for x > 1. Hint: Use the fact that Dlnx = 1/x for x>0.

How does one go about picking an interval to use the Mean Value Theorem? There must be a better strategy than what I'm doing.

 

[MAME]

take derivates of each part?

 

[dvdrdiscs]

Originally posted by: MAME
take derivates of each part?

this is real analysis. we're proving how derivatives work so i can't just take the derivative 😉

 

[miniMUNCH]

Originally posted by: MAME
take derivates of each part?

The Mean Value Theorem is the basis for almost all theorems concerning differentiation including L'Hospital's rules so you can't just use LH rules here. But you can take derivatives and, in fact, the hint is to this effect: use '1/x' for D (ln x).

I think you can prove it using an arbitary interval (a,b) with a=1.

Using generalized form of theorem (dividing f'(c) by g'(c), f and g meet theorems requirements):

[f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c) --> g(b) = g(a) + [g'(c)/f'(c)]*[f(b)-f(a)]

^-- g(a) and f(a) are going to equal zero for functions in question with a=1.

SO: g(b) = [g'(c)/f'(c)]*f(b)

then I think you can show ln x = (x-1)/x * (some function of dervatives of lnx and (x-1)/x). As long as the 'quotient of derivatives' term in above is greater than 1 for all x in [a,b] --> Then g(b) > f(b).

and so forth for x-1...

 

[dvdrdiscs]

Originally posted by: miniMUNCH

Originally posted by: MAME
take derivates of each part?

The Mean Value Theorem is the basis for almost all theorems concerning differentiation including L'Hospital's rules so you can't just use LH rules here. But you can take derivatives and, in fact, the hint is to this effect: use '1/x' for D (ln x).

I think you can prove it using an arbitary interval (a,b) with a=1.

Using generalized form of theorem (dividing f'(c) by g'(c), f and g meet theorems requirements):

[f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c) --> g(b) = g(a) + [g'(c)/f'(c)]*[f(b)-f(a)]

^-- g(a) and f(a) are going to equal zero for functions in question with a=1.

SO: g(b) = [g'(c)/f'(c)]*f(b)

then I think you can show ln x = (x-1)/x * (some function of dervatives of lnx and (x-1)/x). As long as the 'quotient of derivatives' term in above is greater than 1 for all x in [a,b] --> Then g(b) > f(b).

and so forth for x-1...

Hm. That seems a little bit more complicated than what I would expected. The mean value theorem only involves f(b) - f(a) = f'(c)(b-a). where are you getting everything else from?