The interval for which \( f(x)=\displaystyle\frac{x^{2}}{2}+\displaystyle\frac{8}{x^{2}},\ x\ne 0 \) is increasing, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (-2,0)\cup(2,\infty) \)
A function is increasing where \( f'(x)\gt 0 \).
\( \displaystyle f'(x)=x-\frac{16}{x^{3}} \)
\( \displaystyle =\frac{x^{4}-16}{x^{3}} \)
\( \displaystyle =\frac{(x-2)(x+2)(x^{2}+4)}{x^{3}} \)
The factor \( x^{2}+4 \) is always positive, so the sign of \( f' \) is the sign of \( \frac{(x-2)(x+2)}{x^{3}} \). On each interval:
\( x\lt -2:\ f'\lt 0 \)
\( -2\lt x\lt 0:\ f'\gt 0 \)
\( 0\lt x\lt 2:\ f'\lt 0 \)
\( x\gt 2:\ f'\gt 0 \)
Hence \( f \) is increasing on \( (-2,0)\cup(2,\infty) \).
Why the other options are incorrect
Option (1) \( (-\infty,-2) \) — Incorrect. \( f'\lt 0 \) on \( (-\infty,-2) \), so \( f \) is decreasing there.
Option (2) \( (-\infty,-2)\cup(2,\infty) \) — Incorrect. It includes \( (-\infty,-2) \), where \( f \) is decreasing, and leaves out \( (-2,0) \), where \( f \) is increasing.
Option (3) \( (-2,2) \) — Incorrect. It contains \( x=0 \), where \( f \) is not defined, and \( (0,2) \), where \( f'\lt 0 \).