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For the function: \[ f(x)=4 x-6 x^{2 / 3} \] on the interval \( [-1,3] \), The absolute min occurs at \( (-1,-10) \) The absolute max occurs at \( ([?],[~]) \)

Ask by Weber Ayala. in the United States
Mar 07,2025

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The absolute maximum occurs at \( (3, -0.48) \).

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Bonus Knowledge

To find the absolute maximum of the function \( f(x) = 4x - 6x^{2/3} \) on the interval \([-1,3]\), we need to evaluate \( f \) at the critical points and the endpoints. The derivative, \( f'(x) = 4 - 4x^{-1/3} \), can help us find critical points. Setting \( f'(x) = 0 \) leads to \( 4 = 4x^{-1/3} \), so \( x^{1/3} = 1 \), which gives \( x = 1 \). Evaluating \( f \) at the endpoints and the critical point: - At \( x = -1 \): \( f(-1) = 4(-1) - 6(-1)^{2/3} = -4 - 6 = -10 \) - At \( x = 1 \): \( f(1) = 4(1) - 6(1)^{2/3} = 4 - 6 = -2 \) - At \( x = 3 \): \( f(3) = 4(3) - 6(3)^{2/3} = 12 - 6 \cdot \sqrt[3]{9} \approx 12 - 17.1 \approx -5.1 \) Comparing these values: - \( f(-1) = -10 \) - \( f(1) = -2 \) - \( f(3) \approx -5.1 \) The absolute maximum occurs at \( (1, -2) \). So, the final statement is: The absolute max occurs at \( (1, -2) \).

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