Find the absolute extrema of the function g(x) = x/ln(x) on the interval [2, 9]

Answer 1

take the derivative of x/ln and set =0

(x/lnx)' = (lnx -x(1/x)/ln^2(x) = 0

solve for x

(lnx -1)/ln^2x =0

lnx -1 =0

lnx =1

x = about 2.7818

check g(2.7818) and g(2) and g(9), the endpoints

g(2) = 2/ln2 = about 2.89

g(9) = 9/ln9 = about 4.1

g(2.8) = 2.8/ln2.8 =2.7

2.7 is the local and global minimum on the interval 2 to 9