Problem 55 - Calculus Early Transcendentals

Question

a. Analyze \(\lim f(x)\) and \(\lim _{x \rightarrow+\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\) analyze \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\) $$f(x)=\frac{3 x^{4}+3 x^{3}-36 x^{2}}{x^{4}-25 x^{2}+144}$$

Step-by-Step Solution

Answer

2. For which values of x do we find vertical asymptotes? 3. What are the limits of f(x) as x approaches the vertical asymptotes from both sides? Please provide the answers in complete sentences.

Step 1: Find the limit of f(x) as x approaches infinity and negative infinity

We have the function f(x) = \(\frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144}\). To find the limits as x approaches infinity and negative infinity, we can divide both the numerator and the denominator by the highest power of x, which is \(x^4\). Then, f(x) = \(\frac{3 + \frac{3}{x} - \frac{36}{x^2}}{1 - \frac{25}{x^2} + \frac{144}{x^4}}\)

Step 2: Calculate the limits at infinity and negative infinity

Now we calculate the limits as x approaches infinity and negative infinity: \(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3 + \frac{3}{x} - \frac{36}{x^2}}{1 - \frac{25}{x^2} + \frac{144}{x^4}} = \frac{3}{1} = 3\) \(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{3 + \frac{3}{x} - \frac{36}{x^2}}{1 - \frac{25}{x^2} + \frac{144}{x^4}} = \frac{3}{1} = 3\) Since both limits have the same value, the function has a horizontal asymptote at y = 3. b) Vertical asymptotes

Step 1: Find the vertical asymptotes

To find the vertical asymptotes, we need to find the values of x where the denominator of the function is zero. To do this, set the denominator equal to zero and solve for x: \(x^4 - 25x^2 + 144 = 0\) This is a quadratic equation in terms of \(x^2\). Let \(u = x^2\), then the equation becomes: \(u^2 - 25u + 144 = 0\) Factoring, we get: \((u - 9)(u - 16) = 0\) So, \(u = 9\) and \(u = 16\). Reverting back to x: \(x^2 = 9 \Rightarrow x = \pm 3\) \(x^2 = 16 \Rightarrow x = \pm 4\) Hence, we have vertical asymptotes at x = -4, -3, 3, and 4.

Step 2: Analyze the limits as x approaches the vertical asymptotes from both sides

We need to evaluate the following limits for each of the vertical asymptotes: \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\), where a = -4, -3, 3, and 4. We can notice that as x approaches either of these vertical asymptotes from either side, either of the two terms (\(3x^3\) and \(36x^2\)) in the numerator will dominate the polynomial, making the entire fraction go to either negative infinity or positive infinity. Thus, we conclude: For x = -4, -3: \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = -\infty\) For x = 3, 4: \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = \infty\) In summary, the function has a horizontal asymptote at y = 3 and vertical asymptotes at x = -4, -3, 3, and 4.

Key Concepts

Horizontal Asymptote

Horizontal asymptotes describe how a function behaves as the input values become very large or very small. In simple terms, they tell us what output values the function approaches as the input either positively or negatively extends towards infinity. When discussing horizontal asymptotes, the focus is on the ends of the graph. For the function \( f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144} \), we look at how \( f(x) \) behaves as \( x \to \infty \) and \( x \to -\infty \).

To determine the horizontal asymptote of this function, divide both the numerator and the denominator by \( x^4 \), which is the highest power in the denominator. This simplifies the function to \( \frac{3 + \frac{3}{x} - \frac{36}{x^2}}{1 - \frac{25}{x^2} + \frac{144}{x^4}} \). When we calculate the limit as \( x \) approaches positive or negative infinity, \( 3/x \), \( 36/x^2 \), \( 25/x^2 \), and \( 144/x^4 \) all approach zero. Thus, the expression becomes \( \frac{3}{1} = 3 \).

The horizontal asymptote is \( y = 3 \), showing that as \( x \) grows extremely large or small, \( f(x) \) nears the value 3.

Vertical Asymptote

Vertical asymptotes occur where the function tends towards infinity. They indicate the \( x \) values where the function cannot exist -- essentially, points of undefined behavior. For the function \( f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144} \), we seek the \( x \) values where the denominator is equal to zero since division by zero is undefined.

First, set the denominator to zero: \( x^4 - 25x^2 + 144 = 0 \). This equation is quadratic in form, with \( u = x^2 \). Solving \( u^2 - 25u + 144 = 0 \) by factoring gives \((u - 9)(u - 16) = 0\), so \( u = 9 \) and \( u = 16 \). Returning \( u \) back to \( x^2 \), we find \( x^2 = 9 \) giving \( x = \pm 3 \), and \( x^2 = 16 \) giving \( x = \pm 4 \).

Therefore, the vertical asymptotes are at \( x = -4, -3, 3, \) and \( 4 \). At these points, the value of \( f(x) \) approaches infinity from one side and negative infinity from the other, defining a disruption in the function’s graph.

Limits at Infinity

Exploring limits at infinity is about understanding the end behavior of functions. Specifically, it involves identifying what happens to a function, \( f(x) \) as \( x \) becomes extremely large or extremely small. In simple terms, we evaluate \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \) to find values that \( f(x) \) appears to approach.

In the exercise, for \( f(x) = \frac{3x^4 + 3x^3 - 36x^2}{x^4 - 25x^2 + 144} \), we perform this evaluation by simplifying the expression using its highest power in the denominator, resulting in \( \lim_{x \to \infty} f(x) = 3 \) and \( \lim_{x \to -\infty} f(x) = 3 \).

This means that, regardless of the direction we lim towards infinity or negative infinity, the function \( f(x) \) settles towards 3. Thus, understanding limits at infinity is crucial for identifying horizontal asymptotes and predicting a function's behavior in extreme scopes.