Question:
Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).
Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).
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To find the absolute maximum and minimum of a function on a closed interval, first find the critical points by setting the first derivative equal to zero. Then evaluate the function at the critical points and the endpoints of the interval.
Updated On: Jan 13, 2026
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Solution and Explanation
The given function is:
\[
f(x) = 2x^3 - 15x^2 + 36x + 1.
\]
First, find the first derivative of \( f(x) \):
\[
f'(x) = 6x^2 - 30x + 36.
\]
Set \( f'(x) = 0 \) to find the critical points:
\[
6x^2 - 30x + 36 = 0.
\]
\[
x^2 - 5x + 6 = 0.
\]
\[
(x - 2)(x - 3) = 0.
\]
Thus, \( x = 2 \) and \( x = 3 \) are critical points.
Next, evaluate \( f(x) \) at the critical points and the endpoints of the interval \( [1, 5] \):
\[
f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24,
\]
\[
f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29,
\]
\[
f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 10 + 1 = 2,
\]
\[
f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 10 + 1 = 56.
\]
The absolute maximum value is \( f(5) = 56 \) and the absolute minimum value is \( f(1) = 24 \).
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