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Chapter 3: Problem 7

Plot the graph of the following functions: (i) \(y=\frac{\cos x}{\cos 2 x}\) (ii) \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\)

Short Answer

Expert verified
Answer: The vertical asymptotes for the function \(y=\frac{\cos x}{\cos 2 x}\) are at \(x = \frac{(2n + 1) \pi}{4}\) for integer values of \(n\). The vertical asymptotes for the function \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\) are at \(x = -4\) and \(x = 2\).

Step by step solution

01

Identify the domain and range.

The domain of a function is the set of all possible inputs (values of \(x\)) for which the function is defined. For this function, the domain will be all values of \(x\) except the points where the denominator is zero (i.e. \(\cos 2 x = 0\)). We can find these points by setting \(\cos 2x = 0\). The general solution is \(x = \frac{(2n + 1) \pi}{4}\), with \(n\) being an integer. Thus, the domain of the function will be \(x \in \mathbb{R}\) except for \(x = \frac{(2n + 1) \pi}{4}\). The range of the function is the set of all possible output values (\(y\)), which will be all real numbers, since cosine functions have a range between -1 and 1.
02

Identify asymptotes and key points.

Vertical asymptotes occur at the points where the function is not defined. In this case, vertical asymptotes will be at \(x = \frac{(2n + 1) \pi}{4}\) for integer values of \(n\). There are no horizontal asymptotes, as the range of the function is all real numbers.
03

Plot the graph.

With the information we have, let's plot the graph of the function. First, plot the vertical asymptotes at \(x = \frac{(2n + 1) \pi}{4}\) using dashed lines. Then, plot the function using the cosine function as a guide, keeping in mind that the maximum and minimum values of the cosine function are 1 and -1, respectively. Since the function has no horizontal asymptotes, the graph will continue to oscillate around the asymptotes. (ii) \(y=\frac{x^{2}+2 x-3}{x^{2}+2 x-8}\)
04

Identify the domain and range.

The domain of the function is the set of all x-values for which the function is defined. In this case, it will be all values of \(x\) except for the points where the denominator is zero. To find these points, factor the denominator: \(x^{2}+2 x-8 = (x+4)(x-2)\) Thus, the domain will be \(x \in \mathbb{R}\) except for \(x = -4, 2\). Since the function is a ratio of polynomials, the range is also all real numbers.
05

Identify asymptotes and key points.

Vertical asymptotes occur at the points where the function is not defined. In this case, vertical asymptotes will be at \(x = -4\) and \(x = 2\). To find a horizontal asymptote, we look at the degree of the polynomials in the numerator and denominator. Since they are the same, there is a horizontal asymptote at \(y = \frac{\text{coefficient of leading term in the numerator}}{\text{coefficient of leading term in the denominator}} = \frac{1}{1} = 1\).
06

Plot the graph.

With the information we've gathered, let's plot the graph of the function. First, plot the vertical asymptotes at \(x = -4\) and \(x = 2\) using dashed lines. Then, plot the horizontal asymptote at \(y = 1\) using a dashed line as well. Finally, plot the graph of the function using these asymptotes as guides.

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Most popular questions from this chapter

Let \(\mathrm{A}\) and \(\mathrm{B}\) be the points of intersection of the parabola \(y=x^{2}\) and the line \(y=x+2\), and let \(C\) be the point on the parabola where the tangent line is parallel to the graph of \(\mathrm{y}=\mathrm{x}+2 .\) Show that the area of the parabolic segment cut from the parabola by the line four-thirds the area of the triangle \(\mathrm{ABC}\).

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