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

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 $\{x\ne 0\}$. Since $x^4>0$ when $x\ne 0$, the sign of the second derivative $f^{″}$ is determined by the sign of the numerator $2x-6$. Indeed, we can easily see that

$$\begin{array}{rl}2x-6& =2(x-3)>0\text{ for }x>3,\\ 2x-6& =2(x-3)<0\text{ for }x<3.\end{array}$$ This implies that $f^{″}>0$ when $x\in (3,\mathrm{\infty })$ and $f^{″}<0$ when $x\in (-\mathrm{\infty },0)\cup (0,3)$.

Therefore, $f$ is concave up in $(3,\mathrm{\infty });$ while $f$ is concave down in $(-\mathrm{\infty },0)\cup (0,3)$.