How do you find the relative extrema for #f(x) = x - log_4x#? Calculus Graphing with the First Derivative Classifying Critical Points and Extreme Values for a Function 1 Answer A. S. Adikesavan Apr 26, 2016 The minimum of #f(x)# is# f(1/(ln 4))=0.957#, nearly. Explanation: 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.- Answer link Related questions How do you find and classify the critical points of #f(x)=x^3#? How do you find the critical points of a rational function? How do you know how many critical points a function has? How many critical points can a cubic function have? How many critical points can a function have? How many critical points can a quadratic polynomial function have? What is the first step to finding the critical points of a function? How do you find the absolute extreme values of a function on an interval? How do you find the extreme values of the function and where they occur? What is the extreme value of a quadratic function? See all questions in Classifying Critical Points and Extreme Values for a Function Impact of this question 1424 views around the world You can reuse this answer Creative Commons License