How do you find the local extrema for $f (x) = x - ln (x)$ $f(x) = x - ln(x)$ on [0.1,4]?

Question

3119 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-12T18:16:58

Local Minima $= f (1) = 1 .$ $=f(1)=1.$

Fun. $f$ $f$ can not have Local Maxima.

Given fun. $f (x) = x - ln x, x \in [0.1, 4] .$ $f(x)=x-lnx, x in [0.1,4].$

We recall that for local extrema, i.e., maxima/minima, $(i) f'(x)=0, (ii) f''(x)<0$ for maxima, &, $f''(x)>0$ for minima.

Now, $f'(x)=0 rArr 1-1/x=0 rArr x=1 in [0.1,4]$

$f'(x)=1-1/x rArr f''(x)=0-(-1/x^2)=1/x^2 rArr f''(1)=1>0.$

Therefore, f has a local minima at $x=1$ , and it is $f(1)=1-ln1=1-0=1.$

Since, $f''(x)=1/x^2>0, AA x in [0.1,4]$ , f can not have any local maxima.

How do you find the local extrema for f(x) = x - ln(x)f(x)=x−ln(x)f(x) = x - ln(x) on [0.1,4]?