Use the mean value theorem to show that (x-1)/x <= ln(x) <= (x-1) for all x >= 0?

Question

I'm not really sure how the MVT applies to something like that. Can anyone help me understand?

Its only for x>0, which makes sense considering two of the functions arent defined at 0

Answer 1

I emailed you a solution which is a little bit clearer than the poster above. Let me know if you have questions.

Answer 2

I made a mistake formerly

Let´s take the interval 1===>x x>0(strictly)
If we apply the mean value theorem to this interval(suppose x>1)

(ln x- ln1)/(x-1)=1[1+a*(x-1)]where a is a number 0 so ln x/(x-1)<1 and ln x/(x-1)>1/(x) (taking a=0 and a=1
so
1/x< ln x/(x-1)<1 so multiplying by x-1>0

(x-1)/x 1



If x <1
(ln 1- ln x)/(1-x)=1/ (x+a(1-x)) 0
if we put a = 0 -lnx/(1-x)<1/x
and if we put a =1 -lnx/(1-x)>1
so
1<-ln x/(1-x) < 1/x and 1-x <- ln x< (1-x )/x
if we multiply by -1 we must change the inequality

x-1>ln x > (x-1)/x QED
sorry for the former mistake

Answer 3

take f(u)=u^(a million/2) the thought says that it exist a quantity c (0x we prooved a) now for b) merely multiply the two contributors of the inequality a) by potential of x^(a million/2)

Answer 4

Would you integrate each of the quantities?