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

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

At a minimum, the first derivative is zero:

    \(f'(x) = 3x^2 - 3(10 - x)^2\)

    \(f'(x) = 0 \Rightarrow x^2 = (10 - x)^2\)

So \(x = 10 - x\), giving x = 5. (The other case, \(x = -(10 - x)\), gives 0 = −10, which is impossible.)

By the second derivative test:

    \(f''(x) = 6x + 6(10 - x) = 60 \gt 0\)

So f has its minimum at x = 5.

Shortcut: The cubes cancel on expanding, so f is a quadratic:

    \(f(x) = x^3 + 1000 - 300x + 30x^2 - x^3 = 30x^2 - 300x + 1000\)

Its vertex is at

    \( \displaystyle x = -\frac{-300}{2 \times 30} = 5 \)