If $f(x) = 2x^3 - 15x^2 + 36x + 1$, $x \in [1, 5]$, then the absolute minimum value of $f(x)$ is:

The function is $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

To find the absolute minimum value, we need to check the critical points and the endpoints of the interval.

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Find the derivative:

$f'(x) = 6x^2 - 30x + 36$

Set equal to zero:

$6x^2 - 30x + 36 = 0$

$x^2 - 5x + 6 = 0$

$(x - 2)(x - 3) = 0$

The critical points are $x = 2$ and $x = 3$. Both lie within $[1, 5]$.

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Evaluate $f(x)$ at $x = 1$:

$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1$

$f(1) = 2 - 15 + 36 + 1$

$f(1) = 24$

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Evaluate $f(x)$ at $x = 2$:

$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1$

$f(2) = 16 - 60 + 72 + 1$

$f(2) = 29$

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Evaluate $f(x)$ at $x = 3$:

$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1$

$f(3) = 54 - 135 + 108 + 1$

$f(3) = 28$

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Evaluate $f(x)$ at $x = 5$:

$f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1$

$f(5) = 250 - 375 + 180 + 1$

$f(5) = 56$

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Comparing all values: $f(1) = 24$, $f(2) = 29$, $f(3) = 28$, $f(5) = 56$

The absolute minimum value of $f(x)$ on $[1, 5]$ is $24$.