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The minimum value of the function $f(x) = x^3 + (10-x)^3$ occurs at: |
$x=0$ $x=3$ $x=5$ $x=\frac{10}{3}$ |
$x=5$ |
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 $x=5$. |
