Find the absolute maximum and minimum values of a function $f$ given by $f(x) = 2x^{3} - 15x^{2} + 36x + 1$ on the interval $[1, 5]$.

(A) We have $f(x) = 2x^{3} - 15x^{2} + 36x + 1$.
Taking the derivative,we get $f'(x) = 6x^{2} - 30x + 36 = 6(x - 2)(x - 3)$.
Setting $f'(x) = 0$ gives the critical points $x = 2$ and $x = 3$.
We evaluate the function at the critical points and the endpoints of the interval $[1, 5]$:
$f(1) = 2(1)^{3} - 15(1)^{2} + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$.
$f(2) = 2(2)^{3} - 15(2)^{2} + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$.
$f(3) = 2(3)^{3} - 15(3)^{2} + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$.
$f(5) = 2(5)^{3} - 15(5)^{2} + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$.
Comparing these values,the absolute maximum value is $56$ at $x = 5$ and the absolute minimum value is $24$ at $x = 1$.