The absolute minimum of the function $f ( x ) = 8 x ^ { 3 } - 4 x ^ { 2 } + 72 x$ on the interval 0 $\le$ x $\le$ 4 is 0.

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To find the absolute minimum of the function on the given interval, we first find its critical points by taking the derivative and setting it to zero. The derivative of $f(x) = 8x^3 - 4x^2 + 72x$ is $f'(x) = 24x^2 - 8x + 72$. Setting $f'(x) = 0$ and solving for $x$ gives the critical points within the interval. However, evaluating the function at the endpoints of the interval (0 and 4) and at any critical points within the interval will give us the minimum and maximum values. The function value at $x = 0$ is $f(0) = 0$, but without solving for the critical points and comparing values, we cannot conclude that 0 is the absolute minimum of the function on the interval [0, 4]. The function's behavior and the values at critical points and endpoints must be compared to determine the absolute minimum.

The Absolute Minimum of the Function f(x)=8x3−4x2+72xf ( x ) = 8 x ^ { 3 } - 4 x ^ { 2 } + 72 xf(x)=8x3−4x2+72x

The Absolute Minimum of the Function $f ( x ) = 8 x ^ { 3 } - 4 x ^ { 2 } + 72 x$