Question

The false statement(s) is / are (A) If \(\mathrm{P}(\mathrm{x})\) is a polynomial, then the function \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{P}(\mathrm{x})}{\mathrm{x}-1}\) has a vertical asymptote at \(\mathrm{x}=1\). (B) A polynomial function has no vertical asymptote and a rational function has atleast one vertical asymptote. (C) If \(\mathrm{f}(\mathrm{x})\) has a vertical asymptote at \(\mathrm{x}=0\), then \(\mathrm{f}\) is undefined at \(\mathrm{x}=0\). (D) A function can have move than two horizontal asymptotes.

Short answer

A) If P(x) is a polynomial, then the function f(x) = P(x) / (x - 1) has a vertical asymptote at x = 1. B) A polynomial function has no vertical asymptote, and a rational function has at least one vertical asymptote. C) If f(x) has a vertical asymptote at x = 0, then f is undefined at x = 0. D) A function can have more than two horizontal asymptotes. Answer: B

Step-by-step solution

Statement A Analysis

First, let's analyze Statement A: If \(\mathrm{P}(\mathrm{x})\) is a polynomial, then the function \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{P}(\mathrm{x})}{\mathrm{x}-1}\) has a vertical asymptote at \(\mathrm{x}=1\). A rational function is a function of the form \(\frac{P(x)}{Q(x)}\), where P(x) and Q(x) are polynomials. In this case, f(x) is a rational function, and it seems to indicate that a rational function has a vertical asymptote at x=1. A vertical asymptote at x=c occurs when the denominator Q(x) equals zero, i.e., when \(Q(c) = 0\). In this case, the denominator is \(x - 1\), so the vertical asymptote would indeed occur at \(x = 1\).

Statement A Conclusion

Statement A is true because the denominator of the function \(\mathrm{f}(\mathrm{x})\) is equal to zero at \(\mathrm{x}=1\), which results in a vertical asymptote at that point.

Statement B Analysis

Second, let's analyze Statement B: A polynomial function has no vertical asymptote and a rational function has at least one vertical asymptote. A polynomial function is a continuous function without any divisions by zero. Consequently, it cannot have a vertical asymptote. However, the claim that a rational function must have at least one vertical asymptote is not true. For a rational function to have a vertical asymptote, there must exist a value c such that the denominator is zero. But it's possible to have a rational function like \(\frac{x^2 + 4x + 4}{x^2 + 1}\), which does not have any vertical asymptotes because the denominator never equals zero.

Statement B Conclusion

Statement B is false because it is not true that all rational functions must have at least one vertical asymptote.

Statement C Analysis

Third, let's analyze Statement C: If \(\mathrm{f}(\mathrm{x})\) has a vertical asymptote at \(\mathrm{x}=0\), then \(\mathrm{f}\) is undefined at \(\mathrm{x}=0\). A function with a vertical asymptote at x=c indicates that the function approaches infinity or negative infinity as the input approaches the vertical asymptote. A vertical asymptote occurs when the denominator of a rational function equals zero. So, if \(\mathrm{f}(\mathrm{x})\) has a vertical asymptote at \(\mathrm{x}=0\), then the denominator of f(x) must be equal to zero at x=0. In this case, the function is undefined at that point.

Statement C Conclusion

Statement C is true because a vertical asymptote at \(\mathrm{x}=0\) means that the function is undefined at that point.

Statement D Analysis

Finally, let's analyze Statement D: A function can have more than two horizontal asymptotes. In general, a function can have at most two horizontal asymptotes - one as x approaches positive infinity and one as x approaches negative infinity. However, some functions, like the sine and cosine functions, have no horizontal asymptotes at all. Moreover, Piece-wise functions or functions defined over different domains can have more than two horizontal asymptotes.

Statement D Conclusion

Statement D is true because some functions, such as piece-wise functions, can have more than two horizontal asymptotes. In conclusion, the false statement is (B).