From the part (g) of this question, we know that
![{\displaystyle f''(x)={\frac {2x-6}{x^{4}}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/45f7179c9f3ac69f7e15bb90f249b95e41734989)
Also, from the part (a), the domain of
![{\displaystyle f}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61)
is
![{\displaystyle \{x\neq 0\}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9ec10b074f0a267677fb54db6ccff8ba558cea9a)
. Since
![{\textstyle x^{4}>0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f1c6b9de49ef51ca62355556cde37e588195054e)
when
![{\textstyle x\neq 0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/46d10b50ea65d73f0e58f40532a3c17ff861bd4b)
, the sign of the second derivative
![{\displaystyle f''}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fbbdf186092f4353b7630fa8dda903e493cbbdc8)
is determined by the sign of the numerator
![{\displaystyle 2x-6}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fd3ed5aaf7bb63aa67750fa393f7ddeabff8c754)
. Indeed, we can easily see that
![{\displaystyle {\begin{aligned}2x-6&=2(x-3)>0{\text{ for }}x>3,\\2x-6&=2(x-3)<0{\text{ for }}x<3.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/baf018b5823a2d7a7692fd2e6b5a62196f49e528)
This implies that
![{\displaystyle f''>0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c61edc50d8a95c025d41245e3d7617101d3a698a)
when
![{\displaystyle x\in (3,\infty )}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/71a6cfca0e3f55df29e4ca988b403cc9b4a02e35)
and
![{\displaystyle f''<0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/196bf3b908308702f67b85106147c2a6ebf71413)
when
![{\displaystyle x\in (-\infty ,0)\cup (0,3)}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8bd2b8418553a7720061053f63743765e4401165)
.
Therefore,
is concave up in
while
is concave down in
.