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To find the absolute maximum and minimum, we first need to find the critical points of the function within the given interval. This involves finding the derivative of the function and setting it equal to zero.
Given $f(x) = 2x^3 - 15x^2 + 36x + 1$, we find its derivative $f'(x)$. $$f'(x) = 6x^2 - 30x + 36$$
Set $f'(x) = 0$ and solve for $x$: $$6x^2 - 30x + 36 = 0$$ Divide by 6: $$x^2 - 5x + 6 = 0$$ Factor the quadratic equation: $$(x - 2)(x - 3) = 0$$ So, the critical points are $x = 2$ and $x = 3$.
Now, we evaluate the function $f(x)$ at the critical points $x = 2$ and $x = 3$, and at the endpoints of the interval $x = 1$ and $x = 5$. $f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$ $f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$ $f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$ $f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$
Comparing the values of $f(x)$ at these points: $f(1) = 24$ $f(2) = 29$ $f(3) = 28$ $f(5) = 56$ The absolute maximum is 56 at $x = 5$, and the absolute minimum is 24 at $x = 1$.
Final Answer: Absolute maximum: 56 at x=5, Absolute minimum: 24 at x=1