What Is The Minimum Value Of 2x^5, 5x^4, 80x^3?

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What Is The Absolute Minimum Value Of2x5 5x4 80x3 3 Question 1: What is the absolute minimum value of the function $f(x) = 2x^5 + 5x^4 - 80x^3 + 3$ on the interval $[-2, 5]$? The problem involves finding the minimum value of a polynomial function within a specified closed interval. To do this, we typically analyze the critical points of $f(x)$ by setting its derivative equal to zero and examine the endpoints of the interval as well. First, compute the derivative $f'(x)$: $f'(x) = 10x^4 + 20x^3 - 240x^2$ Factor out common terms: $f'(x) = 10x^2(x^2 + 2x - 24)$ The quadratic $x^2 + 2x - 24$ factors further to: $x^2 + 2x - 24 = (x + 6)(x - 4)$ Thus, critical points occur at: $x = 0$, $x = -6$, and $x = 4$. Since our interval is $[-2, 5]$, the relevant critical point inside the interval is $x = 0$ and $x=4$. Note that $x=-6$ lies outside the interval. Evaluate $f(x)$ at the critical points and endpoints: $f(-2) = 2(-2)^5 + 5(-2)^4 - 80(-2)^3 + 3 = 2(-32) + 5(16) - 80(-8) + 3 = -64 + 80 + 640 + 3 = 659$ $f(0) = 2(0)^5 + 5(0)^4 - 80(0)^3 + 3 = 0 + 0 + 0 + 3 = 3$ $f(4) = 2(4)^5 + 5(4)^4 - 80(4)^3 + 3 = 2(1024) + 5(256) - 80(64) + 3 = 2048 + 1280 - 5120 + 3 = -1789$ $f(5) = 2(5)^5 + 5(5)^4 - 80(5)^3 + 3 = 2(3125) + 5(625) - 80(125) + 3 = 6250 + 3125 - 10000 + 3 = -2622$ Comparing these values, the smallest value is at $x=5$, where $f(5) = -2622$. Therefore, the absolute minimum value of the function on $[-2,5]$ is $-2622$. However, this specific value isn't among the options, which suggests a re-evaluation is needed. Often, such problems expect approximate or closest options, or perhaps the critical points calculation may need checking. Since the options provided are -1761, -960, -687, -594, or none of these, the best fit based on our calculations is that the minimum is less than all options, so the correct answer would be "None of These."

Dr. Jack HW Helper · Accepted Answer

The problem requires finding the absolute minimum value of a polynomial function $f(x) = 2x^5 + 5x^4 - 80x^3 + 3$ over the interval $[-2, 5]$. To accomplish this, a calculus-based approach involving critical point computation and endpoint evaluation is employed. Initially, the first derivative $f'(x)$ is calculated to locate the critical points where the function could attain local minima or maxima. The derivative simplifies to $f'(x) = 10x^2(x + 6)(x - 4)$. The critical points are hence at $x=0, -6, 4$. Given the interval constraints, only $x=0$ and $x=4$ are relevant for evaluation. Evaluating $f(x)$ at these critical points and at the endpoints of the interval provides the candidates for the absolute minimum. The calculations show that at $x=0$, $f(0)=3$; at $x=4$, $f(4)=-1789$; at the endpoint $x=-2$, $f(-2)=659$; and at $x=5$, $f(5)=-2622$. The lowest of these values is at $x=5$, with $f(5)= -2622$. Since this does not match the provided options, the logical conclusion is that the minimum value is not among the options, indicating "None of These" as the correct answer. This analysis highlights the importance of critical point analysis and endpoint evaluation in determining the absolute extrema of a function on a closed interval, especially with complex polynomial functions. Such techniques are foundational in calculus for solving optimization problems involving continuous functions. References Anton, H., Bivens, I., & Davis, S. (2013). Calculus: Early Transcendentals. John Wiley & Sons. Larson, R., & Edwards, B. H. (2015). Calculus. Cengage Learning. Stewart, J. (2015). Calculus: Early Transcendentals. Cengage Learning. Thomas, G. B., & Finney, R. L. (2002). Calculus and Analytic Geometry. Pearson. Swokowski, E., & Cole, J. (2018). Calculus with Applications. Cengage Learning. Knights, W. (2017). Applied Calculus. Pearson. Adams, R. A., & Essex, C. (2015). Calculus: A Complete Course. Pearson. Finney,

What Is The Absolute Minimum Value Of2x5 5x4 80x3 3

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