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Given that f(–5) = f(3) = 0 and f '(-1) = 0, and f '(x) < 0 for x < -1 and f '(x) > 0 for x > -1, what can be concluded about the function f(x)?

User SubRed
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Final answer:

The function f(x) has zeros at x = -5 and x = 3, a relative minimum at x = -1 due to a horizontal tangent and a change from decreasing to increasing at that point.

Step-by-step explanation:

Based on the information provided, f(-5) = f(3) = 0 implies that the function f(x) has zeros at x = -5 and x = 3. The given f '(-1) = 0 suggests that there is a horizontal tangent to the graph of f(x) at x = -1. Additionally, since f '(x) < 0 for x < -1 and f '(x) > 0 for x > -1, it can be inferred that f(x) is decreasing for x values less than -1 and increasing for x values greater than -1.

Put together, these facts suggest that f(x) has a relative minimum at x = -1. This is because the derivative changes from negative to positive, indicating a change from a decreasing to an increasing function. Consequently, the point (-1, f(-1)) must be a local minimum.

User Sonny Prince
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