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Chapter 9: Problem 37

Sketch the graph of the function. Label the intercepts, relative extrema, points of inflection, and asymptotes. Then state the domain of the function. \(y=\frac{2 x}{x^{2}-1}\)

Short Answer

Expert verified
The function \(y=\frac{2 x}{x^{2}-1}\) doesn't have any intercepts, extrema or inflection points. The function has vertical asymptotes at \(x= \pm 1\) and a horizontal asymptote at \(y = 0\). The domain is all real numbers except \(x = -1\) and \(x = 1\).

Step by step solution

01

Find the intercepts

To find the y-intercept, set \(x = 0\) in the equation. To find the x-intercept, set \(y = 0\) in the equation. In this case, when setting \(y = 0\), we find that the equation \(0 = \frac{2x}{x^2 - 1}\) implies that \(x = 0\), however x=0 is not in the domain of the function, hence there are no x-intercepts. When setting \(x = 0\) in the equation, we get a division by zero - again, not in the domain of the function; so, there is no y-intercept.
02

Find the asymptotes

Vertical asymptotes occur where the denominator of the fraction equals zero (assuming the numerator does not also equal zero). This equation is \(x^{2}-1=0\), which gives the solutions \(x= \pm 1\). The horizontal asymptote is found by looking at the degree of the polynomial in the numerator and denominator. Because the degree of the denominator is greater than the degree of the numerator, the function will approach 0 as \(x \rightarrow \pm \infty\). Thus, the horizontal asymptote is \(y = 0\).
03

Identify extrema and points of inflection

First, find the derivative of the function, \(y' = - \frac{2x^2 - 2}{\left( x^2 - 1 \right)^2}\). The critical points will be where \(y'=0\) or undefined. However, in this case \(y'\) is never zero and is undefined for \(x^2 - 1 = 0\), namely \(x = \pm 1\). These points are not in the domain of the original function and thus can't be extrema or inflection points. Then identify possible points of inflection by looking at the second derivative, \(y'' = \frac{12x}{\left( x^2 - 1 \right)^3}\), which simplifies to \(0 = \frac{12x}{\left( x^2 - 1 \right)^3}\) gives \(x = 0\), but this point will not bring about a sign change in the second derivative and hence there are no inflection points.
04

Find the domain

The domain of a function is the values of \(x\) for which the function is defined. In this case, the function is undefined for any \(x\) such that the denominator equals zero, \(x^{2}-1=0\). Hence, the domain of this function is \(-\infty < x < -1\) union \(1 < x < \infty\)

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