Science:Math Exam Resources/Courses/MATH100/December 2016/Question 09 (c) (i)

MATH100 December 2016

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Question 09 (c) (i)
Let $f(x)={\frac {\sqrt {x^{2}-8}}{x-4}}$. Note that ${\begin{aligned}f''(x)&={\frac {8x^{2}(x-3)}{(x^{2}-8)^{3/2}(x-4)^{3}}}.\end{aligned}}$ (i) Find all intervals where $f(x)$ is concave down.

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Hint
In order to determine the concavity of a given function at each point, we need to look at the sign of the second derivative of the function at that point.

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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. First, we find the points at which $f''$ is zero or undefined within the domain $\left((-\infty ,-2{\sqrt {2}}]\cup [2{\sqrt {2}},\infty )\right)\setminus \{4\}$ of $f$. We are given that $f''(x)={\frac {8x^{2}(x-3)}{(x^{2}-8)^{3/2}(x-4)^{3}}}$, so the only such points are $x=3$ (where the numerator is zero) and $x=\pm 2{\sqrt {2}},\,4$ (where the denominator is zero). We use the following sign chart to determine the sign of $f''$. $\displaystyle (-\infty ,-2{\sqrt {2}})$ $(2{\sqrt {2}},3)$ $\displaystyle (3,4)$ $(4,\infty )$ $8x^{2}$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $+$ $(x-3)$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $+$ $(x^{2}-8)^{3/2}$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $+$ $(x-4)^{3}$ $\displaystyle -$ $\displaystyle -$ $-$ $\displaystyle +$ $f''(x)$ $\displaystyle +$ $\displaystyle +$ $\displaystyle -$ $+$ $f(x)$ concave up concave up concave down concave up This implies that the function is concave down on the interval $\color {blue}(3,4).$