0 votes 0 votes minimum of the real valued function f(x)=(x-1)^2/3 occurs at x equal to ?? how to find the value of when equating f’(x)=0??? eyeamgj asked Dec 13, 2018 eyeamgj 1.5k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Prateek Raghuvanshi commented Dec 13, 2018 i edited by Prateek Raghuvanshi Dec 13, 2018 reply Follow Share actually it is not differential at x=1 ,we can write this, $(\sqrt[3]{(x-1)})^2$ we can see at x=1 ,it will give 0 which is minimum value . 0 votes 0 votes eyeamgj commented Dec 13, 2018 reply Follow Share ok but there will be square on over cuberoot?? 0 votes 0 votes Prateek Raghuvanshi commented Dec 13, 2018 reply Follow Share it is corrected now , thank you 0 votes 0 votes eyeamgj commented Dec 13, 2018 reply Follow Share thank u for help 1 votes 1 votes Please log in or register to add a comment.