Science:Math Exam Resources/Courses/MATH100/December 2016/Question 09 (b) (ii)

MATH100 December 2016

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Question 09 (b) (ii)
Let $f(x)={\frac {\sqrt {x^{2}-8}}{x-4}}$. (ii) Find all intervals where $f(x)$ 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
$f(x)$ is increasing on $(a,b)$ if $f'(x)>0$ on the interval. Be careful that the function must be defined at the intervals that you choose for monotonicity.

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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. From part (i), we have $f'(x)={\frac {\sqrt {x^{2}-8}}{x-4}}\left({\frac {x}{x^{2}-8}}-{\frac {1}{x-4}}\right)={\frac {\sqrt {x^{2}-8}}{x-4}}\left({\frac {x(x-4)-(x^{2}-8)}{(x^{2}-8)(x-4)}}\right)={\frac {4{\sqrt {x^{2}-8}}(2-x)}{(x^{2}-8)(x-4)^{2}}}$. Note that the function $f$ is defined when $x^{2}-8\geq 0$ and $x\neq 4$ and its derivative is defined when $x^{2}-8>0$ and $x\neq 4$. This implies that on the domain of $f'$, we have $${\frac {4{\sqrt {x^{2}-8}}}{(x^{2}-8)(x-4)^{2}}}>0,$$ so the sign of $f'(x)$ is determined by the sign of $2-x$. Indeed, we have $f'(x)>0$ when $x<2,$ and $f'(x)<0$ when $x>2.$ However, we note that $f$ is defined on $(-\infty ,-2{\sqrt {2}}]\cup [2{\sqrt {2}},4)\cup (4,\infty ),$ and $f'$ is defined on $(-\infty ,-2{\sqrt {2}})\cup (2{\sqrt {2}},4)\cap (4,\infty ).$ Therefore, $f$ is increasing on $\color {blue}(-\infty ,-2{\sqrt {2}})$.