The provided function is:
\[
f(x) = 2x^3 - 15x^2 + 36x + 1.
\]
The first derivative of \( f(x) \) is calculated as:
\[
f'(x) = 6x^2 - 30x + 36.
\]
To find the critical points, we set \( f'(x) = 0 \):
\[
6x^2 - 30x + 36 = 0.
\]
Dividing by 6 yields:
\[
x^2 - 5x + 6 = 0.
\]
Factoring the quadratic equation gives:
\[
(x - 2)(x - 3) = 0.
\]
Therefore, the critical points are \( x = 2 \) and \( x = 3 \).
Next, we evaluate \( f(x) \) at the critical points and the interval endpoints \( [1, 5] \):
\[
f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24,
\]
\[
f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29,
\]
\[
f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 10 + 1 = 2,
\]
\[
f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 10 + 1 = 56.
\]
The absolute maximum value is \( f(5) = 56 \), and the absolute minimum value is \( f(1) = 24 \).