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Consider the function f(x)=5(x-3)² / ³. For this function there are two important intervals: (-, A) and (A, ) where A is a critical number. Find A For each

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Final answer:

The critical number A for the quadratic function f(x)=5(x-3)^²/3 is 3, which divides the domain into two intervals: (-∞, 3) and (3, ∞). In continuous probability functions, probabilities correspond to areas under the curve of the function.

Step-by-step explanation:

The original function presented in the question, f(x)=5(x-3)^²/3, is a quadratic function. The critical number A in this context is the value of x where the derivative of f(x) is zero, indicating a potential maximum or minimum of the function. Since we have a squared term, the graph will have a parabolic shape opening upwards, with the vertex representing the critical number. For the standard quadratic function f(x) = a(x - h)^² + k, the vertex is at the point (h, k). In our case, the vertex is at (x = 3, y = 0), making the critical number A=3. Therefore, we have two intervals: (-∞, 3) and (3, ∞).

For the SEO keywords 'continuous probability function' and 'PROBABILITY = AREA' referenced in the question, these relate to the properties of probability density functions in which the total area under the curve and above the x-axis equals 1. In such functions, the probability of an event within a certain range is found by integrating the function over that range. For example, if we have a continuous probability function f(x) defined from 0 to 7, the question P(x = 10) is extraneous as it falls outside the defined range. Similarly, for a continuous distribution P(x = 7) would technically be 0 since a single point has no area under a continuous curve.

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