- Identify that the function must have a non-zero denominator for all real \(x\) to be continuous everywhere.
- Compute the limit as \(x \to \infty\) for each function to find the horizontal asymptote \(y = 1\).
- Option \(h(x) = \frac{x^2-16}{x^2+16}\) is the only function satisfying both criteria: it is defined for all \(x\) and its horizontal asymptote is \(y = 1\).
- The correct function is \(\boxed{h(x) = \frac{x^2-16}{x^2+16}}\).
Explanation
- Analysis of Requirements
We are tasked with identifying which of the four given rational functions is continuous for all real numbers and possesses a horizontal asymptote at \(y = 1\). A rational function is defined as the ratio of two polynomials. Continuity for all real numbers implies that the denominator must have no real roots. A horizontal asymptote at \(y = L\) is determined by the limit of the function as \(x\) approaches infinity or negative infinity, \(\lim_{x \to \infty} h(x) = L\).
- Evaluating the Options
Let us evaluate each option:
- \(h(x) = \frac{x^2-16}{x^2+16}\): The denominator \(x^2 + 16\) is never zero for any real \(x\), meaning this function is continuous for all real numbers. Evaluating the limit as \(x \to \infty\): \(\lim_{x \to \infty} \frac{x^2-16}{x^2+16} = \lim_{x \to \infty} \frac{1 - 16/x^2}{1 + 16/x^2} = \frac{1-0}{1+0} = 1\). This function satisfies both conditions.
- \(h(x) = \frac{x^2+16}{x^2-16}\): The denominator is zero when \(x^2 = 16\), i.e., \(x = \pm 4\). Thus, the function has vertical asymptotes at \(x = 4\) and \(x = -4\), violating the continuity requirement.
- \(h(x) = \frac{x^2-16}{x-4}\): The function is undefined at \(x = 4\), meaning it is not continuous everywhere. Furthermore, \(\lim_{x \to \infty} \frac{x^2-16}{x-4} = \lim_{x \to \infty} (x+4) = \infty\), so it does not have a horizontal asymptote at \(y=1\).
- \(h(x) = \frac{x+4}{x^2+16}\): The denominator is never zero, so it is continuous everywhere. However, \(\lim_{x \to \infty} \frac{x+4}{x^2+16} = \lim_{x \to \infty} \frac{1/x + 4/x^2}{1 + 16/x^2} = 0\). Thus, the horizontal asymptote is at \(y=0\), not \(y=1\).
- Conclusion
Based on the evaluation, the only function that is continuous for all real numbers and has a horizontal asymptote at \(y=1\) is \(h(x) = \frac{x^2-16}{x^2+16}\).
Examples
In electronics, rational functions are frequently used to model transfer functions of filters. For example, a low-pass filter's gain might be modeled by a rational function. Knowing if the function is continuous ensures the circuit doesn't have "gaps" where it fails to respond, and the horizontal asymptote tells us the steady-state gain of the filter at high frequencies. Ensuring these properties helps engineers design stable and predictable communication systems.