Prove that 1/x+lnx is never less than 1

Answer 1

f(x) = 1/x + lnx

take the derivative and set =0 to find the minimum point

f'(x) = -1/x^2 + 1/x = 0

1/x = 1/x^2

multiply both sides by x^2 to get

x = 1

f(1) = y=1/1+ln1 = 1

(1, f(1)) = (1,1) is graphically the minimum point