Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (h)

MATH110 April 2016

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Question 05 (h)
Let $f(x)={\frac {x-1}{x^{2}}}$. (h) Find the intervals where $f$ is concave up and the intervals where it is concave down.

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Hint
Recall that (1) ${\textstyle f}$ is concave up (convex) on the intervals where ${\textstyle f''(x)>0}$; (2) ${\textstyle f}$ is concave down on the intervals where ${\textstyle f''(x)<0}$;

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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 the part (g) of this question, we know that $$f''(x)={\frac {2x-6}{x^{4}}}.$$ Also, from the part (a), the domain of $f$ is $\{x\neq 0\}$. Since ${\textstyle x^{4}>0}$ when ${\textstyle x\neq 0}$, the sign of the second derivative $f''$ is determined by the sign of the numerator $2x-6$. Indeed, we can easily see that $${\begin{aligned}2x-6&=2(x-3)>0{\text{ for }}x>3,\\2x-6&=2(x-3)<0{\text{ for }}x<3.\end{aligned}}$$ This implies that $f''>0$ when $x\in (3,\infty )$ and $f''<0$ when $x\in (-\infty ,0)\cup (0,3)$. Therefore, ${\textstyle f}$ is concave up in ${\textstyle \color {blue}(3,\infty );}$ while ${\textstyle f}$ is concave down in ${\textstyle \color {blue}(-\infty ,0)\cup (0,3)}$.