How do you find the relative extrema for #f(x) = x - log_4x#?

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Answer 1 · 2016-04-26T15:08:20

The minimum of #f(x)# is# f(1/(ln 4))=0.957#, nearly.

Use #log_b a = log_c a log_a c#

#ln x=log_e x=log_4 x log_e 4=log_4 x ln 4#.

#log_4 x=ln x/ln 4#.

Now, #f(x)=x-ln x/ln 4#

#f'=1-1/(x ln 4)=0#, when #x=1/ln 4#

#f''=1/(x^2 ln 4)>0#, for all x, as #ln 4=1.3863>0#..

Thus,# f(1/ln 4)# is the minimum of f(x).

#f(1/ln 4)=1/ln 4-ln(1/ln 4)/ln 4=(1+ln(ln 4))/ln 4=0.957#, nearly.-