\(f(x) = 2x^3 - 15x^2 + 36x + 1\) implies
\[f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)\]\[f'(x) = 6x^2 - 30x + 36\]Set \(f'(x) = 0\) to find critical points:\[6x^2 - 30x + 36 = 0\]
\[\implies x^2 - 5x + 6 = 0\]\[\implies(x - 2)(x - 3) = 0\implies x=2, 3\]The critical points are \(x=2\) and \(x=3\).
Note that both the critical points are les in the interval \([1, 5]\)
Now evaluate \(f(x)\) at the critical points that lie in the interval and at the endpoints.
For (\(x=1\)) we have \(f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1\)=\(24\)
For (\(x=2\)) we have \(f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1\)=\(29\)
For (\(x=3\)) we have \(f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1\)=\(28\)
For (\(x=5\)) we have \(f(5) = 2(125) - 15(25) + 36(5) + 1 = 250 - 375 + 180 + 1\)=\(56\)
Hence Absolute Maximum Value: \(\mathbf{56}\) (Occurs at \(x=5\)) and Absolute Minimum Value: \(\mathbf{24}\) (Occurs at \(x=1\))