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

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Answer 1 · 2016-07-12T18:16:58

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

Fun. #f# can not have Local Maxima.

Given fun. #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.