2024年12月11日 · Consequently, this function has a relative maximum, not a minimum. Function represented by the table of points: The y-values in the table decrease to -6 at x = 0 and then increase, indicating a minimum point at (0, -6). This shows that there is a vertex (minimum) for the quadratic represented by these points, confirming it opens upwards.

2024年12月4日 · Determine the Relative Minimum: A relative minimum is a point where the function value is less than the values of the function at nearby points. To find the relative minimum, we usually look for critical points by setting the first derivative, f ′ (x) = 0. For the function f (x) = x 3 − 2 x 2 − 4, the derivative is f ′ (x) = 3 x 2 − 4 x.

2022年12月23日 · The function h(x) has a relative minimum at x = 2, determined by setting the first derivative of h(x) equal to zero and confirming it with the sign change in the second derivative. Explanation: The question revolves around finding where the function h(x), defined as the integral from -5 to x of a given function, has a relative minimum.

2025年1月16日 · If f ′′ (x) > 0 at a critical point, the function has a relative minimum there. If f ′′ (x) < 0 at a critical point, the function has a relative maximum there. Approximated Results. Using graphical or numerical methods, we approximate: Real Zeros: The approximate real zeros of the function are x ≈ − 1.8 and x ≈ 2.

2024年12月15日 · Therefore, the critical point at x = − 8 1 is indeed a local maximum, not a minimum. Conclusion. After analyzing the function and its derivatives, we conclude that f (x) = 9 x 2/3 + 3 x − 6 has no relative minimum at the given options. Hence the correct choice regarding the relative minimum is: No relative minimum in the provided options.

2023年3月28日 · Therefore, by the second partials test, we can conclude that the critical point (-4, 3) is a relative minimum of the function. h(x, y) = x^2 − 7xy − y^2 To find the critical points, we need to find where the partial derivatives of h with respect to x and y are both equal to zero:

2025年1月2日 · However, the statement that there must always be a relative minimum between two relative maxima is false. Consider the function f(x) = -x^4 + 4x^2. It has two relative maxima at x = -1 and x = 1, but there is no relative minimum between them, as the function decreases without creating a local minimum in that interval. Conclusion

2024年4月5日 · To find the relative minimum of , we need to find the critical points of the function, which are the values of x where the derivative of is equal to . Taking the derivative of with respect to , we get: Setting , we get: Evaluating at , we get: = However, the search results indicate that h(x) does not have a relative minimum at , but rather at and .

2024年11月26日 · To determine the relative minimum of the function f (x) = 9 x 2/3 + 3 x − 6, we can go through the following steps: Find the First Derivative: The first step in identifying relative minimums is to find the derivative of f (x) with respect to x. The derivative, denoted as f ′ (x), is found using the power rule and standard differentiation ...

2021年3月2日 · To determine the relative minimum from the given options, we need to understand what a relative minimum is. A relative minimum is the lowest point in a small region of the graph. In simpler terms, if you picture the graph as a landscape, the relative minimum is like a valley, where the point is lower than the surrounding points on both sides ...