Question

Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Short answer

Answer: To provide a short answer based on the solution, we can state: The domain of the given function is \(x\in(-\infty,-6)\cup(-6,6)\cup(6,+\infty)\). To find the areas of \(R_1\) and \(R_2\), we need to set up and evaluate the corresponding integrals: $$ A(R_1) = \int_{-12}^{-\frac{12}{\sqrt 3}} \frac{1}{-x\sqrt{x^{2}-36}} dx $$ $$ A(R_2) = \int_{\frac{12}{\sqrt 3}}^{12} \frac{1}{x\sqrt{x^{2}-36}} dx $$ These integrals can be calculated by using integration techniques such as substitution or partial fractions or with the help of a calculator.

Step-by-step solution

Find the domain of \(f(x)\)

The function is given by: \(f(x)=\frac{1}{x\sqrt{x^2-36}}\). We have a fraction, so we need to determine the values of \(x\) for which the denominator is not equal to zero. Denominator: \(x\sqrt{x^2-36}\). We must have \(x\neq 0\) and \(x^2-36\neq 0\). For the latter, we have \(x^2\neq 36\) or \(x\neq \pm 6\). So, the domain is \(x\in(-\infty,-6)\cup(-6,6)\cup(6,+\infty)\).

Graph \(f(x)\) on its domain

To graph the function, we can identify the end behavior and asymptotes: - As \(x\to \pm\infty\), \(f(x)\to 0\). - Vertical asymptotes occur when \(x=\pm 6\). - Horizontal asymptote happens at \(y=0\). Use graphing calculator or software to create the graph of the function.

Calculate the areas of \(R_1\) and \(R_2\)

First, we partition \(R_1\): \(R_1: \text{ between } -12 \text{ and } -\frac{12}{\sqrt 3}\) and \(R_2\): \(R_2: \text{ between } \frac{12}{\sqrt 3} \text{ and } 12\) To find the area of \(R_1\), we can set up an integral over the interval \([-12,-\frac{12}{\sqrt 3}]\): $$ A(R_1) = \int_{-12}^{-\frac{12}{\sqrt 3}} |f(x)| dx = \int_{-12}^{-\frac{12}{\sqrt 3}} \frac{1}{-x\sqrt{x^{2}-36}} dx $$ To find the area of \(R_2\), we set up an integral over the interval \([\frac{12}{\sqrt 3}, 12]\): $$ A(R_2) = \int_{\frac{12}{\sqrt 3}}^{12} |f(x)| dx = \int_{\frac{12}{\sqrt 3}}^{12} \frac{1}{x\sqrt{x^{2}-36}} dx $$ Next, we can calculate these integrals either by hand (using substitution or partial fractions) or with a calculator. After calculating the integrals, compare the areas with the graph to ensure that they are consistent.

Key concepts

Graphing Functions

Graphing functions is a critical skill in calculus that helps visualize how a function behaves across its domain. Understanding the graph of a function gives insight into its continuity, limits, and asymptotic behavior. To graph the function \(f(x)=\frac{1}{x\sqrt{x^2-36}}\), first pinpoint key features such as intercepts and asymptotes:
\- The function lacks intercepts because there are no values of \(x\) for which both the numerator is non-zero and the denominator equates to zero in this case.
\- Next, identify asymptotes to understand the function's behavior near the boundaries where it cannot be evaluated. To effectively graph this function, make use of graphing calculators or software to plot the curve. This provides a visual confirmation of mathematical calculations and helps in accurately capturing the behavior of the function.

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable. For the function \(f(x)=\frac{1}{x\sqrt{x^2-36}}\), the expression in the denominator dictates the domain limitations. Here’s how to find the domain of this function:
\- Set the denominator \(x\sqrt{x^2-36}\) not equal to zero.
\- \(xeq0\) because division by zero is undefined.
\- Also, \(x^2-36eq0\) implies \(xeq\pm 6\). Thus, the function is defined for all \(x\) except at these specific points. Therefore, the domain is \(x\in(-\infty,-6)\cup(-6,6)\cup(6,\infty)\). Recognize how the domain impacts the intervals over which the function is graphed.

Definite Integrals

Definite integrals are used to calculate the area under a curve between two points on the x-axis. They are a fundamental tool in calculus for measuring cumulative quantities. For our function, the task is to find the areas \(R_1\) and \(R_2\) given by the integral:
\- For \(R_1\), integrate over \([-12,-\frac{12}{\sqrt 3}]\).
\- For \(R_2\), integrate over \([\frac{12}{\sqrt 3}, 12]\). The expression becomes:
\[ A(R_1) = \int_{-12}^{-\frac{12}{\sqrt 3}} \left| \frac{1}{x\sqrt{x^2-36}} \right| \, dx \]
\[ A(R_2) = \int_{\frac{12}{\sqrt 3}}^{12} \left| \frac{1}{x\sqrt{x^2-36}} \right| \, dx \] Integrate carefully as improper integrals may arise near vertical asymptotes. Use analytical or numerical methods, like substitution or calculators, for the calculations. The results will provide the bounded area, reflective of the behavior shown in the graph.

Asymptotes

Asymptotes are lines that a curve approaches as it heads towards infinity. They provide crucial insights into the behavior of functions and help in sketching accurate graphs. For \(f(x)=\frac{1}{x\sqrt{x^2-36}}\), the asymptotes are:
\- Horizontal asymptote: As \(x \to \pm \infty\), \(f(x) \to 0\). This indicates the function levels off to zero at extreme values of \(x\).
\- Vertical asymptotes occur at \(x=\pm 6\). The function is undefined at these points, and the graph approaches these lines without ever touching them.Understanding asymptotes is crucial for accurately graphing the function and sketching its limits of behavior. It guides the plotting process and ensures a comprehensive depiction of the function across its domain.