At what points are the functions in Exercises 13–30 continuous? y = 1/(|x| + 1) − x²/2
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Step 1: Understand the concept of continuity. A function is continuous at a point if the limit of the function as it approaches the point from both sides is equal to the function's value at that point.
Step 2: Break down the function y = 1/(|x| + 1) − x²/2 into its components. The function consists of two parts: 1/(|x| + 1) and −x²/2.
Step 3: Analyze the continuity of the first part, 1/(|x| + 1). The expression |x| represents the absolute value of x, which is continuous everywhere. The denominator |x| + 1 is never zero, so 1/(|x| + 1) is continuous for all x.
Step 4: Analyze the continuity of the second part, −x²/2. The function x² is a polynomial, and polynomials are continuous everywhere. Therefore, −x²/2 is continuous for all x.
Step 5: Combine the results from steps 3 and 4. Since both components of the function are continuous for all x, the entire function y = 1/(|x| + 1) − x²/2 is continuous for all x.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Continuity of Functions
A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. This means there are no breaks, jumps, or holes in the graph of the function at that point. For a function to be continuous over an interval, it must be continuous at every point within that interval.
Piecewise functions are defined by different expressions based on the input value. Understanding how these functions behave at the boundaries of their defined pieces is crucial for determining continuity. In the given function, the absolute value and polynomial components may create different behaviors depending on the value of x.
Limits are fundamental in calculus for analyzing the behavior of functions as they approach specific points. To determine continuity, one must evaluate the limit of the function as x approaches a point from both the left and right. If these limits exist and are equal to the function's value at that point, the function is continuous there.