the function is increasing when its derivative is greater than 0, so take the derivative of the function, we have f'(x)= (1-lnx)/x^2>0, or 1>lnx, xNext, the function is concaved upward only when the second derivative is positive, so we take the second derivative of the function, and get f''(x)=(-x-(1-lnx)2x)/x^4>0, or x+(1-lnx)2x<0, which is x(1+2(1-lnx))=x(3-2lnx)<0. So they have different signs, and since x can't be negative, we have x>0 and 3-2lnx<0, which is lnx>1.5, x>e^1.5.
