Question:
If \( f(x) = 2x^3 - 3x^2 + 4 \), then the minimum value of \( f(x) \) in the interval \( [0, 3] \) is:
If \( f(x) = 2x^3 - 3x^2 + 4 \), then the minimum value of \( f(x) \) in the interval \( [0, 3] \) is:
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To find the minimum value of a function in a closed interval, compute the derivative, find critical points, and evaluate the function at critical points and endpoints.
Updated On: May 24, 2025
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Solution and Explanation
To find the minimum value of \( f(x) = 2x^3 - 3x^2 + 4 \) in the interval \( [0, 3] \), we use the following steps:
- Compute the first derivative to find critical points: \[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 4) = 6x^2 - 6x = 6x(x - 1) \]
- Set \( f'(x) = 0 \): \[ 6x(x - 1) = 0 \implies x = 0 \text{ or } x = 1 \]
- Both \( x = 0 \) and \( x = 1 \) lie in \( [0, 3] \).
- Evaluate \( f(x) \) at critical points and endpoints (\( x = 0, 1, 3 \)):
- At \( x = 0 \): \( f(0) = 2(0)^3 - 3(0)^2 + 4 = 4 \)
- At \( x = 1 \): \( f(1) = 2(1)^3 - 3(1)^2 + 4 = 2 - 3 + 4 = 3 \)
- At \( x = 3 \): \( f(3) = 2(3)^3 - 3(3)^2 + 4 = 54 - 27 + 4 = 31 \)
- Compute the first derivative to find critical points: \[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 4) = 6x^2 - 6x = 6x(x - 1) \]
- Set \( f'(x) = 0 \): \[ 6x(x - 1) = 0 \implies x = 0 \text{ or } x = 1 \]
- Both \( x = 0 \) and \( x = 1 \) lie in \( [0, 3] \).
- Evaluate \( f(x) \) at critical points and endpoints (\( x = 0, 1, 3 \)):
- At \( x = 0 \): \( f(0) = 2(0)^3 - 3(0)^2 + 4 = 4 \)
- At \( x = 1 \): \( f(1) = 2(1)^3 - 3(1)^2 + 4 = 2 - 3 + 4 = 3 \)
- At \( x = 3 \): \( f(3) = 2(3)^3 - 3(3)^2 + 4 = 54 - 27 + 4 = 31 \)
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