Find the inverse of the function f(x) = (2x - 1)/(x + 3)

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To find the inverse of the function f of x equals 2x minus 1 divided by x plus 3, we'll follow a step-by-step process. First, we replace f of x with y to get y equals 2x minus 1 divided by x plus 3. Next, we swap the variables x and y, which gives us x equals 2y minus 1 divided by y plus 3. This is the key step in finding the inverse function. On the right, you can see the graph of our original function, which has a vertical asymptote at x equals negative 3 and a horizontal asymptote at y equals 2. Now we need to solve for y to find the inverse function. Starting with x equals 2y minus 1 divided by y plus 3, we multiply both sides by y plus 3 to get x times y plus 3 equals 2y minus 1. Distributing x, we get xy plus 3x equals 2y minus 1. Rearranging to isolate terms with y, we get xy minus 2y equals negative 3x minus 1. Factoring out y gives us y times x minus 2 equals negative 3x minus 1. Finally, dividing both sides by x minus 2, we get y equals negative 3x minus 1 divided by x minus 2. This is our inverse function. On the graph, the blue curve represents our original function, while the red curve shows its inverse. Notice how they're reflections of each other across the line y equals x, which is a key property of inverse functions. Now that we've found the inverse function f inverse of x equals negative 3x minus 1 divided by x minus 2, let's verify it's correct. A key property of inverse functions is that f of f inverse of x equals x. Let's substitute our inverse function into the original function. We get f of f inverse of x equals f of negative 3x minus 1 divided by x minus 2. Substituting this into our original function formula, we get a complex fraction that we need to simplify. After several algebraic steps, including multiplying fractions and combining like terms, we eventually get negative 7x divided by negative 7, which equals x. This confirms that our inverse function is correct. Note that we can also write the inverse function in an alternative form as 3x plus 1 divided by 2 minus x, which is obtained by multiplying both numerator and denominator by negative 1. Let's analyze the domain and range of both our original function and its inverse. For the original function f of x equals 2x minus 1 divided by x plus 3, the domain is all real numbers except x equals negative 3, where the denominator becomes zero. The range is all real numbers except y equals 2, which is the horizontal asymptote. For the inverse function f inverse of x equals negative 3x minus 1 divided by x minus 2, the domain is all real numbers except x equals 2, and the range is all real numbers except y equals negative 3. Notice the relationship between these sets: the domain of the inverse function equals the range of the original function, and the range of the inverse function equals the domain of the original function. This is a fundamental property of inverse functions. On the graph, we can see the vertical asymptotes at x equals negative 3 for the original function in blue, and at x equals 2 for the inverse function in red. Similarly, the horizontal asymptotes are at y equals 2 for the original function and y equals negative 3 for the inverse function. Let's summarize what we've learned about finding the inverse of a rational function. To find the inverse of a function, we follow these steps: replace f of x with y, swap the variables x and y, and then solve for y. For our specific function f of x equals 2x minus 1 divided by x plus 3, we found that the inverse function is f inverse of x equals negative 3x minus 1 divided by x minus 2. We verified this result by checking that f of f inverse of x equals x. We also observed important relationships between domains and ranges: the domain of the inverse function equals the range of the original function, and the range of the inverse function equals the domain of the original function. Graphically, we saw that inverse functions are reflections of each other across the line y equals x. This process works for many types of functions, though the algebraic steps may vary in complexity depending on the original function.