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The minimum value of the function $f(x)=x^2+\frac{128}{x}$ is : |
64 48 112 96 |
48 |
The correct answer is Option (2) → 48 $f(x)=x^2+\frac{128}{x}$ $f'(x)=2x-\frac{128}{x^2}$ for critical points, $f'(c)=0$ $⇒2c-\frac{128}{c^2}=0$ $⇒2c^3=128$ $⇒c^3=64$ $⇒c=4$ $f(4)=16+\frac{128}{4}=48$ |
