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

MATH104 December 2011

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Question 01 (g)
Find the interval(s) on which $\displaystyle f(x)=x^{2}-\ln(x)$ is increasing.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A function is increasing when its slope is positive.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\left({\frac {1}{\sqrt {2}}},\infty \right)$

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