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Chapter Review, Question Problem 95AYU

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The given expression is Use the law of exponents The principal square root is nonnegative. The absolute value is .
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Problem 3: Determining the Domain and Graph of a Function Given the function f(x) = √(x^2 - 4), we need to determine the domain and graph of the function. To find the domain, we need to consider the values of x that make the expression inside the square root non-negative. In other words, we need to find the values of x for which x^2 - 4 ≥ 0. Solving the inequality, we have: x^2 - 4 ≥ 0 (x - 2)(x + 2) ≥ 0 The critical points are x = -2 and x = 2. We can create a sign chart to determine the intervals where the expression is non-negative: -2 2 |-----|-----| - + From the sign chart, we can see that the expression is non-negative when x ≤ -2 or x ≥ 2. Therefore, the domain of the function is (-∞, -2] ∪ [2, ∞). Next, let's graph the function. We can start by plotting the critical points (-2, 0) and (2, 0). Since the function is a square root, it will only be defined for x-values that make the expression inside the square root non-negative. For x ≤ -2, the function becomes f(x) = √(x^2 - 4). Plugging in a value such as x = -3, we get f(-3) = √((-3)^2 - 4) = √(9 - 4) = √5. So, we can plot the point (-3, √5) on the graph. Similarly, for x ≥ 2, the function becomes f(x) = √(x^2 - 4). Plugging in a value such as x = 3, we get f(3) = √((3)^2 - 4) = √(9 - 4) = √5. So, we can plot the point (3, √5) on the graph. Connecting the points and considering the asymptote at x = 2 and x = -2, we get the graph of the function.
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Question Text
Chapter Review, Question Problem 95AYU
TopicAll Topics
SubjectPre Calculus
ClassClass 11
Answer TypeText solution:1