The minimum value of the function f(x) = x^3 + (10-x)^3 occurs at:

The correct answer is Option (3) → $x=5$

$f(x)=x^{3}+(10-x)^{3}$

$f'(x)=3x^{2}-3(10-x)^{2}$

$3x^{2}-3(100-20x+x^{2})=0$

$3x^{2}-300+60x-3x^{2}=0$

$60x-300=0$

$x=5$

$f''(x)=6x+6(10-x)=60 > 0$

Minimum value occurs at $x=5$.