We are studying the rational function f of x equals 2 x over x minus 1. A rational function is defined as the ratio of two polynomials. In this case, the numerator is 2x and the denominator is x minus 1.
To find the domain of our function, we must identify values that make the denominator equal to zero, since division by zero is undefined. Setting the denominator x minus 1 equal to zero gives us x equals 1. Therefore, the domain is all real numbers except x equals 1, written as the union of intervals from negative infinity to 1 and from 1 to positive infinity.
Asymptotes help us understand the behavior of the function at extreme values. The vertical asymptote occurs where the function is undefined, at x equals 1. For horizontal asymptotes, we compare the degrees of the numerator and denominator. Since both are degree 1, we divide the leading coefficients: 2 divided by 1 equals 2. So the horizontal asymptote is y equals 2.
Let's examine how the function behaves near its asymptotes and at extreme values. As x approaches 1 from the left, the function approaches negative infinity. As x approaches 1 from the right, the function approaches positive infinity. As x approaches positive or negative infinity, the function approaches the horizontal asymptote y equals 2.
Finding intercepts and key points helps us plot the function accurately on a graph. To find the y-intercept, we evaluate f at 0, which gives us 0. So the y-intercept is at point (0, 0). For other key points: f of 2 equals 4, giving us point (2, 4). f of negative 1 equals 1, giving us point (-1, 1).
Now we'll combine all our findings to sketch the complete graph of f of x. We have a vertical asymptote at x equals 1 and a horizontal asymptote at y equals 2. The function has a y-intercept at (0, 0). We've calculated additional points: (2, 4), (-1, 1), and (3, 3). The graph consists of two separate branches - one to the left of the vertical asymptote and one to the right. Both branches approach but never touch the asymptotes.
Rational functions appear in real-world scenarios like concentration problems. Let's solve an example: Find x when f of x equals 3. We set up the equation 2x over x minus 1 equals 3. Multiplying both sides by x minus 1 gives us 2x equals 3 times x minus 1. Expanding the right side: 2x equals 3x minus 3. Subtracting 3x from both sides: negative x equals negative 3. Therefore, x equals 3. We can verify this by substituting back into the original function: f of 3 equals 6 over 2, which equals 3.