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Graph \( g(x)=-x^{4}+3 x \). Identify the \( x \)-intercepts and the points where the local maximums and local minimums occur. Determi which the function is increasing or decreasing. Round to the nearest hundredth, if necessary. From least to greatest, the \( x \)-intercepts are \( x=\square \) and \( x \approx \square \). The local maximum is \( (\square, \square \) ). The function is increasing when \( x<\square \). The function is decreasing when \( x>\square \).

Ask by Garrett Montgomery. in the United States
Feb 03,2025

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From least to greatest, the \( x \)-intercepts are \( x = 0 \) and \( x \approx 1.44 \). The local maximum is \( (0.91, 2.05) \). The function is increasing when \( x < 0.91 \). The function is decreasing when \( x > 0.91 \).

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To analyze the function \( g(x) = -x^4 + 3x \), we first find the \( x \)-intercepts by setting \( g(x) = 0 \). Factoring gives us \( -x(x^3 - 3) = 0 \), leading to \( x = 0 \) and \( x \approx \sqrt[3]{3} \approx 1.44 \). From least to greatest, the \( x \)-intercepts are \( x = 0 \) and \( x \approx 1.44 \). Next, we find the local maxima and minima by calculating the derivative \( g'(x) = -4x^3 + 3 \) and setting it to zero. This gives us critical points. Solving \( -4x^3 + 3 = 0 \) results in \( x \approx \sqrt[3]{\frac{3}{4}} \approx 0.91 \) for a local maximum. Evaluating \( g(0.91) \) gives the local maximum point. Upon further investigation, the function \( g(x) \) is increasing on the interval \( (-\infty, 0.91) \) and decreasing on the interval \( (0.91, \infty) \). Therefore, the local maximum is approximately \( (0.91, g(0.91)) \) with \( g(0.91) \) computed for the final output. To summarize: - The local maximum is found at roughly \( (0.91, 2.73) \). - The function increases when \( x < 0.91 \) and decreases when \( x > 0.91 \).

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