The minimum value of \(f(x) = x^{2} + \displaystyle\frac{16}{x^{2}}\), \(x \ne 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8
Setting the derivative equal to zero:
\( \displaystyle f'(x) = 2x - \frac{32}{x^{3}} = 0 \)
\( x^{4} = 16 \), so \(x = \pm 2\)
The second derivative is
\( \displaystyle f''(x) = 2 + \frac{96}{x^{4}} \gt 0 \)
so both points are points of local minimum, with the same value:
\( \displaystyle f(\pm 2) = 4 + \frac{16}{4} = 8 \)
These are the only critical points, and f(x) becomes very large as x approaches 0 or as |x| grows, so 8 is the minimum value.