Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (g)/Solution 1

The given function is increasing where its first derivative is positive. Taking the derivative yields

$\displaystyle f'(x)=2x-1/x$

We must now solve the inequality for x yields the interval where ƒ is increasing. We will only be interested in positive values of x, as negative values of x are not in the domain of the given function (because of the natural logarithm). Therefore,

${\begin{aligned}0&<2x-1/x,\quad x>0\\1/x&<2x\\1&<2x^{2}\\{\frac {1}{2}}&<x^{2}\\\end{aligned}}$

Therefore, we have

$x>+{\frac {1}{\sqrt {2}}}$

The negative solution was ruled out because of the requirement that x be a positive number.

The interval on which ƒ is increasing is

$\left({\frac {1}{\sqrt {2}}},\infty \right)$