Differentiating exponentials-Help me thanks, would be a great help to me!

[Unusualskill]

(a) find stationary value of y=lnx - x, and deduce that ln x<=x-1 and x>0 with equality only when x=1
(b)Find the stationary value of ln x+ (1/x) , and deduce that (x-1)/x <=ln x for x>0 with equality only when x=1
(c) by putting x= z/y, where 0<y<z, deduce Napier's inequality , 1/z < {(ln z-ln y)/(z-y )} <1/y

 

[Unusualskill]

what have you tried so far?

do you know what \(\displaystyle \dfrac{d}{dx}\ln(x)\) equals?

ya i know, im just wondering of the deduce part...thanks

 

[Unusualskill]

ok. What is the stationary point of \(\displaystyle \ln(x)-x\) ?

Is it a maximum or minimum?

stationary value =-1 when x=1 . it is a maximum value with d2y/dx2=-1

 

[Unusualskill]

ok good.

That tells you that \(\displaystyle \ln(x)-x\leq-1\) for all x > 0.

Doing a bit of rearranging that gets you

\(\displaystyle \ln(x)\leq (x-1)\) for all x > 0

see if you can tackle b and c now.

thanks alot!

 

[pka]

deduce Napier's inequality , 1/z < {(ln z-ln y)/(z-y )} <1/y.

It may interest you to see how this usually proven. Use the mean value theorem.
If \(\displaystyle f\) is differentiable on \(\displaystyle [a,b]\) then \(\displaystyle (\exists c \in (a,b))\left[ {f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}} \right]\).

In this case \(\displaystyle f(x)=\ln(x)\) and \(\displaystyle f'(x)=\dfrac{1}{x}\) and \(\displaystyle 0<y<c<z\).
That gives \(\displaystyle \dfrac{1}{z} < \dfrac{{f(z) - f(y)}}{{z - y}} = f'(c) = \dfrac{1}{c} < \dfrac{1}{y}\)