Question

Let f: R-->R
Use the Mean Value Theorem to show that (x-1)/x
< log x < (x-1), for x > 1

Answer 1

for derivatives

Answer 2

for example x-1>log x let f(x)=x-1-log x\[f'(x)=1-\frac{1}{x}>0 \ \ \ \text{for all} \ \ x>1\]

Answer 3

f(x) is continuous and differentiable on \((1, \infty)\) so we can apply MVT

Answer 4

so if x>1 \[\frac{f(x)-f(1)}{x-1}=f'(c)\]for some c>1 ..... f(1)=0\[f(x)=(x-1)f'(c)>0 \]since x-1>0 and f'(c)>0

Answer 5

it implies that \[x-1-\log x>0 \ \ \ \text{for all} \ \ x>1\]or\[x-1>\log x\]do the same proccess for other part

Answer 6

thanks, lol i treated logx having a base of 10 xD. so that's why i had a hard time proving it. thanks again.