Science:Math Exam Resources/Courses/MATH104/December 2016/Question 11 (c)

MATH104 December 2016

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Question 11 (c)
Consider the function $f(x)={\frac {x^{2}-2}{(x-2)^{2}}}$, and its derivatives $$f'(x)={\frac {4(1-x)}{(x-2)^{3}}},~~~~~f''(x)={\frac {4(2x\color {red}-\color {black}1)}{(x-2)^{4}}}.$$ Note: The expression for the second derivative given in the exam PDF is incorrect; the correct expression is given here. (c) Determine where ${\textstyle f(x)}$ is concave up or down and find the inflection points.

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Hint
A function $f$ is concave up(down) on an interval $(a,b)$ if $f''(x)>0(<0)$ on it.

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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. Note that $f$ is concave up(down) on an interval $(a,b)$ if $f''(x)>0(<0)$ on it. Therefore it is enough to determine the sign of the second derivative of $f$; $f''(x)={\frac {4(2x-1)}{(x-2)^{4}}}$ Since the denominator is always positive on the domain, the sign of $f''$ follows its numerator. Using $2x-1>0$ on $\left({\frac {1}{2}},\infty \right)$ and $2x-1<0$ on $\left(-\infty ,{\frac {1}{2}}\right)$, $f''>0$ and hence $f$ is concave up on ${\textstyle \left({\frac {1}{2}},2\right),(2,\infty )}$ while $f''<0$ and hence$f$ is concave down on ${\textstyle \left(-\infty ,{\frac {1}{2}}\right)}$. Here $x=2$ is excluded because it is not in the domain of $f$. Finally, $f''$ vanishes at $x={\frac {1}{2}}$ and $f''$ changes its sign around the point, $\left({\frac {1}{2}},f\left({\frac {1}{2}}\right)\right)=\left({\frac {1}{2}},-{\frac {7}{9}}\right)$ is an inflection point of $f$. Here, $f\left({\frac {1}{2}}\right)$ is obtained from $f\left({\frac {1}{2}}\right)={\frac {(1/2)^{2}-2}{(3/2)^{2}}}={\frac {-7/4}{9/4}}={\frac {-7}{9}}$. To sum, $\color {blue}{\text{CCU on }}\left({\frac {1}{2}},2\right),(2,\infty ),{\text{CCD on }}\left(-\infty ,{\frac {1}{2}}\right),{\text{inflection point}}\left({\frac {1}{2}},-{\frac {7}{9}}\right)$