Problem 36 - Precalculus: Mathematics for Calculus

Question

A Function from a Description Find a formula from a function \(f\) that satisfies the following conditions. $$\begin{aligned} &\lim _{x \rightarrow \infty} f(x)=0, \quad \lim _{x \rightarrow 0} f(x)=-\infty, \quad f(2)=0\\\ &\lim _{x \rightarrow 3^{-}} f(x)=\infty, \quad \lim _{x \rightarrow 3^{+}} f(x)=-\infty \end{aligned}$$

Step-by-Step Solution

Answer

The function \( f(x) = \frac{A(x-2)}{(x-3)^2} \) satisfies the given conditions.

Step 1: Interpret the Limits

The condition \( \lim_{x \rightarrow \infty} f(x) = 0 \) suggests that \( f(x) \) approaches 0 as \( x \) becomes very large. Similarly, \( \lim_{x \rightarrow 0} f(x) = -\infty \) implies that as \( x \) approaches 0 from the positive side, the function tends toward negative infinity.

Step 2: Identify the Function Form

The given limits \( \lim_{x \rightarrow 3^{-}} f(x)=\infty \) and \( \lim_{x \rightarrow 3^{+}} f(x)=-\infty \) suggest a vertical asymptote at \( x = 3 \). Additionally, \( f(2) = 0 \) suggests that 2 is a root of the function. An appropriate form of \( f(x) \) would be a rational function such as \( \frac{a}{b-cx} \) adjusted to fit exact conditions.

Step 3: Construct a Function

Based on the above observations, a form for \( f(x) \) that includes a vertical asymptote at \( x = 3 \), a root at \( x = 2 \), and approaches zero as \( x \rightarrow \infty \) could be \( f(x) = \frac{A(x-2)}{-(x-3)} \). This function meets all the given conditions except optionally requiring a manipulation to meet horizontal asymptote at zero as \( x \rightarrow \infty \).

Step 4: Adjust the Function's Parameters

To ensure \( \lim_{x \rightarrow \infty} f(x) = 0 \), the degree of the numerator should be less than the degree of the denominator. Thus, \( f(x) = \frac{A(x-2)}{(x-3)^2} \). The extra \( (x-3) \) in the denominator provides an additional degree, ensuring the horizontal asymptote condition is satisfied.

Step 5: Verification and Conclusion

Verify that the constructed function \( \frac{A(x-2)}{(x-3)^2} \) satisfies the limit conditions. \( \lim_{x \rightarrow \infty} \frac{A(x-2)}{(x-3)^2} = 0 \) holds, \( \lim_{x \rightarrow 3^{-}} \frac{A(x-2)}{(x-3)^2} = \infty \), and \( \lim_{x \rightarrow 3^{+}} \frac{A(x-2)}{(x-3)^2} = -\infty \) also hold due to behavior near \( x = 3 \). \( f(2) = 0 \) as needed. Solving for \( A \) is not necessary unless specific range conditions alter.

Key Concepts

Limits

In precalculus, limits help us understand the behavior of a function as the input nears a certain value, either a number or infinity. When we say \( \lim_{x \to \infty} f(x) = 0 \), it means that as \( x \) grows larger and larger, the output of the function \( f(x) \) approaches zero. It's like telling you what's happening way down the line, toward infinity.

When discussing \( \lim_{x \to 0} f(x) = -\infty \), we're looking at what \( f(x) \) does as \( x \) gets closer and closer to zero from the positive side. Here, the function dives down towards negative infinity. Limits are crucial for predicting a function's behavior near critical points without knowing the exact value at those points. They're like giving a heads-up for extremes or approaching behaviors without a full explanation necessary at each step.

Asymptotes

Asymptotes are lines that a graph approaches but never actually touches or crosses. There are two main types: vertical and horizontal asymptotes.

Vertical asymptotes occur where a function takes the form of (like this problem). They show that as the function approaches a certain \( x \) value, such as \( x = 3 \) in our exercise, the output shoots up to positive infinity from the left \( \lim_{x \rightarrow 3^{-}} \) or dives to negative infinity from the right \( \lim_{x \rightarrow 3^{+}} \).

Horizontal asymptotes are slightly different. They show the behavior of a function as \( x \) moves towards positive or negative infinity. In our case, \( \lim_{x \rightarrow \infty} f(x) = 0 \) indicates that as \( x \) increases, \( f(x) \) levels off towards 0, suggesting a horizontal asymptote at \( y = 0 \).

Understanding these lines is important for graph interpretation and predicting function behavior in various scenarios. Think of asymptotes as sort of invisible guidelines, telling us where a graph might aim as we move through its domain.

Rational Functions

Rational functions are a type of mathematical expression where one polynomial is divided by another. These are crucial in understanding complex functions as they often depict real-world scenarios where inputs affect outputs in nonlinear ways.

In this exercise, we've fashioned a rational function \( f(x) = \frac{A(x-2)}{(x-3)^2} \). This specific form helps satisfy the given conditions of limits and asymptotic behavior. Here, the numerator \( A(x-2) \) features a root at \( x = 2 \), because when \( x \) equals 2, the entire function equals zero.

Meanwhile, the denominator \( (x-3)^2 \) has a more complex role. It produces a vertical asymptote at \( x = 3 \) where the function becomes undefined. The square in the denominator also helps in controlling the horizontal asymptote at 0 as \( x \) approaches infinity, achieving a balance in polynomial degrees. In rational functions, the degree of the numerator and denominator deeply influences asymptotic behavior, so choosing them wisely helps meet the conditions you're solving for. Rational functions are versatile and powerful, essential tools in modeling and problem-solving in precalculus.