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

MATH110 April 2016

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Question 05 (e)
Let $f(x)={\frac {x-1}{x^{2}}}.$ (e) Find the intervals where $f$ is increasing and the intervals where it is decreasing.

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Hint
Calculate the derivative, and determine the intervals where ${\textstyle f'(x)>0}$ and ${\textstyle f'(x)<0}$. When ${\textstyle f'(x)>0}$, ${\textstyle f(x)}$ is increasing, while ${\textstyle f'(x)<0}$, ${\textstyle f(x)}$ is decreasing.

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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. Let’s calculate the derivative first by the quotient rule, $$f'(x)={\frac {(x-1)'x^{2}-(x^{2})'(x-1)}{x^{4}}}={\frac {x^{2}-2x^{2}+2x}{x^{4}}}={\frac {(2-x)x}{x^{4}}}={\frac {2-x}{x^{3}}}.$$ Recall that the domain of $f$ in part (a) is $\{x\neq 0\}$. On the domain, the only solution of $f'(x)=0$ is ${\textstyle 2}$. Therefore, at the points $0{\text{ and }}2$, the derivative ${\textstyle f'(x)}$ might change its sign. So, we make a partition of intervals in the real line based on these points and examine the sign of ${\textstyle f'(x)}$; (1) when ${\textstyle x\in (-\infty ,0)}$, ${\textstyle x^{3}<0,\ 2-x>0}$, so ${\textstyle f'(x)<0}$; (2) when ${\textstyle x\in (0,2)}$, ${\textstyle x^{3}>0,\ 2-x>0}$, so ${\textstyle f'(x)>0}$; (3) when ${\textstyle x\in (2,\infty )}$, ${\textstyle x^{3}>0,\ 2-x<0}$, so ${\textstyle f'(x)<0}$. By the Hint, this tells us that ${\textstyle f}$ is increasing in the interval ${\textstyle \color {blue}(0,2)}$, while it is decreasing in ${\textstyle \color {blue}(-\infty ,0)\cup (2,\infty ).}$