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

Let’s calculate the derivative first by the quotient rule, $$f^{\prime }(x)=\frac{(x-1{)}^{\prime }{x}^2-(x^2{)}^{\prime }(x-1)}{x^4}=\frac{x^2-2x^2+2x}{x^4}=\frac{(2-x)x}{x^4}=\frac{2-x}{x^3}.$$

Recall that the domain of $f$ in part (a) is $\{x\ne 0\}$. On the domain, the only solution of $f^{\prime }(x)=0$ is $2$.

Therefore, at the points $0\text{ and }2$, the derivative $f^{\prime }(x)$ might change its sign. So, we make a partition of intervals in the real line based on these points and examine the sign of $f^{\prime }(x)$;

(1) when $x\in (-\mathrm{\infty },0)$, $x^3<0,2-x>0$, so $f^{\prime }(x)<0$;

(2) when $x\in (0,2)$, $x^3>0,2-x>0$, so $f^{\prime }(x)>0$;

(3) when $x\in (2,\mathrm{\infty })$, $x^3>0,2-x<0$, so $f^{\prime }(x)<0$.

By the Hint, this tells us that $f$ is increasing in the interval $(0,2)$, while it is decreasing in $(-\mathrm{\infty },0)\cup (2,\mathrm{\infty }).$