Question

Graphing with technology Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points. $$f(x)=\frac{3 x-5}{x^{2}-1}$$

Short answer

Question: List the key features used to graph the function $$f(x)=\frac{3 x-5}{x^{2}-1}$$ and their respective values. Answer: The key features used to graph the function are: 1. Domain: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\) 2. Intercepts: x-intercept at \((\frac{5}{3}, 0)\) and y-intercept at \((0, 5)\) 3. Asymptotes: Vertical asymptotes at x = 1 and x = -1, and horizontal asymptote at y = 0 4. Critical points: Potential extreme points at x ≈ 0.383 and x ≈ 2.617

Step-by-step solution

Find the Domain

To find the domain, we look for values of x for which the function is defined. In this case, the function is not defined when the denominator is equal to 0. Find the points where the denominator equals 0. \(x^2-1=0\) \((x-1)(x+1)=0\) Therefore, x = 1 and x = -1 are the points where the function is not defined. The domain of the function is: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\)

Find the Intercepts

To find the x-intercept, set f(x) = 0 and solve for x. $$\frac{3 x-5}{x^{2}-1} = 0$$ Since the numerator has to equal 0 for the function to equal 0, $$3x - 5 = 0$$ $$x = \frac{5}{3}$$ Thus, the x-intercept is at the point \((\frac{5}{3}, 0)\) To find the y-intercept, set x = 0 and solve for f(x). $$f(0) = \frac{3(0)-5}{0^{2}-1} = \frac{-5}{-1} = 5$$ The y-intercept is at the point \((0,5)\).

Find the Asymptotes

For vertical asymptotes, we'll consider the points where the function is not defined. In this case, we have two vertical asymptotes at x = 1 and x = -1. For the horizontal asymptote, we'll find the limit of the function as x approaches infinity. $$\lim_{x\to\infty} \frac{3 x-5}{x^{2}-1} = \lim_{x\to\infty} \frac{3-\frac{5}{x}}{x-\frac{1}{x}} = 0$$ Thus, we have a horizontal asymptote at y = 0.

Find the Critical Points

To find local extreme values and inflection points, we'll first need to find the first and second derivative of the function and then determine where they meet necessary conditions. The first derivative of the function: $$f'(x) = \frac{d}{dx} \frac{3x-5}{x^2 -1} = \frac{-3x^2 + 5x - 3 + 6x}{(x^2-1)^2} = \frac{-3x^2 + 11x - 3 }{(x^2-1)^2}$$ To find potential local extreme points, set the first derivative equal to 0 and solve for x. $$-3x^2 + 11x - 3 = 0$$ \(x \approx 0.383, 2.617\) Now we will check where the second derivative f''(x) changes concavity for inflection points. The second derivative of the function: $$f''(x) = \frac{d^2}{dx^2} \frac{-3x^2 + 11x - 3 }{(x^2-1)^2} = \frac{(6x-11)(x^2-1)^2 - 2(x^2-1)(-6x^2+22x - 6)}{((x^2-1)^2)^2.}$$ There are no points where the second derivative is equal to 0 and within the domain of the original function. Now using these key features, you can use a graphing utility to plot the function and its key points: 1. Domain: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\) 2. Intercepts: x-intercept at \((\frac{5}{3}, 0)\) and y-intercept at \((0, 5)\) 3. Asymptotes: Vertical asymptotes at x = 1 and x = -1, and horizontal asymptote at y = 0 4. Critical points: Potential extreme points at x ≈ 0.383 and x ≈ 2.617 These features will help you create a complete graph of the function $$f(x) = \frac{3x-5}{x^2 -1}$$.

Key concepts

Finding the Domain of a Function

Understanding the domain of a function is fundamental as it represents all possible input values (typically the x-values) for which the function is defined. When graphing rational functions, we must be particularly attentive to values that could cause the denominator to be zero, as a function is undefined when its denominator is zero. For the rational function \(f(x)=\frac{3x-5}{x^2-1}\), we set the denominator equal to zero and solve for x: \(x^2-1=0\). Factoring gives us \(x-1)(x+1)=0\), leading to two values where the function is undefined: x = 1 and x = -1. With this, we can express the domain as \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty)\). It is helpful to visualize this on a number line to clearly illustrate which sections encompass the domain of our function.

Locating Intercepts

Intercepts offer a quick way to find points where the graph of a function crosses the x-axis and y-axis. For the x-intercept, the y-value is zero, and for the y-intercept, the x-value is zero.
To find the x-intercept of our function, \(f(x)\), we set \(f(x)=0\) and solve for x. In this equation, \(\frac{3x-5}{x^2-1} = 0\), the numerator must be zero, which gives us the x-intercept at \(x = \frac{5}{3}\). Similarly, setting x to zero gives us the y-intercept, which for \(f(x)\) is \(f(0) = 5\). Thus, the graph will cross the x-axis at \(\left(\frac{5}{3}, 0\right)\) and the y-axis at \(\left(0, 5\right)\). It is often useful to mark these intercepts on a graph as they are key points that help shape the curve of the function.

Determining Asymptotes

Asymptotes are lines that the graph of a function approaches but never actually reaches. Vertical asymptotes occur at undefined points of a function, and horizontal asymptotes reflect the behavior as x goes to infinity or negative infinity.
In the case of \(f(x)=\frac{3x-5}{x^2-1}\), vertical asymptotes appear at the points where the denominator is zero—x = -1 and x = 1. To find the horizontal asymptote, we examine the limits as x approaches infinity and as x approaches negative infinity. Here, both limits yield a value of 0, indicating a horizontal asymptote at y = 0. Graphing these asymptotes helps us understand the behavior of the function and alerts us to where the function will have dramatic increases or decreases in value.

Calculating Derivatives for Critical Points

Critical points of a function offer insights into its local maximums, minimums, and points of inflection. To locate these points on a graph, we calculate the first and second derivatives of the function. These derivatives help us determine where the function is increasing or decreasing and where its concavity changes.
The first derivative of our function, \(f'(x)\), tells us where the function might have potential local extreme points by setting it equal to zero. For \(f'(x)\) of \(f(x)=\frac{3x-5}{x^2-1}\), the critical points are approximately at x ≈ 0.383 and x ≈ 2.617. The second derivative, \(f''(x)\), helps in finding points of inflection, which are points where the graph changes concavity. However, for our function, there are no points where \(f''(x)=0\) within the domain, meaning there are no points of inflection. These calculations are key to creating a more accurate graph and understanding the overall shape of the function.