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ID:
Class: 12
Subject: Math
Topic: Application of Derivative
Type: Long
Question:
Find the maximum value of \(2x^3 - 24x + 107\) in the interval \([1,3]\). Find the maximum value of the same function in \([-3, -1]\).
Official Solution
Explanation:
Let
\[
f(x) = 2x^3 - 24x + 107
\]
Differentiate:
\[
f'(x) = 6x^2 - 24 = 6(x^2 - 4)
\]
Set \( f'(x) = 0 \):
\[
x^2 - 4 = 0 \Rightarrow x = \pm 2
\]
For the Interval \([1, 3]\):
Evaluate the function at:
\(x = 1\):
\[
f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85
\]
and at \(x = 2\):
\[
f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75
\]
At \(x = 3\):
\[
f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89
\]
\[
\boxed{\text{Maximum value in } [1,3] \text{ is } 89 \text{ at } x = 3}
\]
For the Interval \([-3, -1]\):
Evaluate the function at:
\(x = -3\):
\[
f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125
\]
and at \(x = -2\):
\[
f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139
\]
and at \(x = -1\):
\[
f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129
\]
\[
\boxed{\text{Maximum value in } [-3, -1] \text{ is } 139 \text{ at } x = -2}
\]
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