The local minimum value of \( f(x)=x^{3}+\displaystyle\frac{243}{x} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 108
\( \displaystyle f'(x)=3x^2-\frac{243}{x^2} \)
At a turning point f'(x) = 0:
\( \displaystyle 3x^4=243 \)
\( \displaystyle x^4=81 \)
\( \displaystyle x=\pm 3 \)
By the second derivative test:
\( \displaystyle f''(x)=6x+\frac{486}{x^3} \)
\( \displaystyle f''(3)=18+18=36\gt 0 \)
\( \displaystyle f''(-3)=-18-18=-36\lt 0 \)
So f has a local minimum at x = 3 (and a local maximum at x = −3).
\( \displaystyle f(3)=27+\frac{243}{3} \)
\( \displaystyle =27+81 \)
\( \displaystyle =108 \)