Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (e)/Solution 1

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 ${\textstyle \color {blue}(-\infty ,0)\cup (2,\infty ).}$