Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) \(f(x)=\sqrt{\frac{1}{x+1}}\) (b) \(g(x)=\sqrt{\frac{x}{x+1}}\)

Step-by-Step Solution

Answer

The domain of \(f(x)\) and \(g(x)\) are all real numbers except -1. For \(f(x)\), x is also not equal to zero. Furthermore, \(f(x)\) increases while \(g(x)\) decreases as we approach -1 from either side. Both of the functions converge to 0 as x approaches negative infinity and positive infinity. And, \(g(x)\) converges to 1 as x approaches positive infinity.

Step 1: Understanding Function \(f(x)=\sqrt{\frac{1}{x+1}}\)

Firstly, please compute the domain of f(x). This function will be defined for all real numbers except for -1. The denominator must be different from zero, so x+1≠0 and x≠-1. Then, analyze the behavior of \(f(x)\) for \(x \rightarrow +\infty\) and \(x \rightarrow -\infty\), except -1.

Step 2: Understanding Function \(g(x)=\sqrt{\frac{x}{x+1}}\)

Similar to \(f(x)\), first compute the domain of \(g(x)\). This function will be defined for all real numbers except for -1. Again, we cannot have a zero denominator so x+1≠0 hence, x≠-1. Also, it should be noted that \(x >= 0\) because it is under a square root sign. Analyze the behavior of \(g(x)\) for \(x \rightarrow +\infty\), \(x \rightarrow -\infty\), except -1, and \(x -> 0\).

Step 3: Behaviors

Examine behaviors around x=-1. Also, observe that function \(f(x)\rightarrow +\infty\) when \(x→-1+\) (approaching -1 from the right), and \(f(x)\rightarrow +\infty\) when \(x→-1−\), (approaching -1 from the left). On the other hand, \(g(x)\rightarrow +\infty\) when \(x→-1+\) (approaching -1 from the right) and \(g(x)\rightarrow 0\) when \(x→-1−\) (approaching -1 from the left). The progressions are \(f(x)\) converges to 0 as \(x\) approaches both \(+\infty\) and \(-\infty\). For \(g(x)\), it converges to 1 as \(x\) approaches \(+\infty\), and it converges to 0 as \(x\) approaches \(0+\) or from the right of 0.

Key Concepts

Limits and behavior analysis

In calculus, the concept of limits helps us understand the behavior of functions as they approach certain points or tend towards infinity. Understanding this allows us to make predictions about the function's output without explicitly computing it everywhere.

For instance, consider the function \(f(x) = \sqrt{\frac{1}{x+1}}\). We need to analyze how this function behaves as \(x\) approaches notable points like -1, infinity, and negative infinity. The limit as \(x\to-1^+\) or \(x\to-1^-\) (from the right or left of -1) for this function approaches infinity because the denominator approaches zero, leading the whole expression under the square root towards infinity.

Similarly, when \(x\) goes towards positive or negative infinity, the function \(f(x)\) approaches zero. This occurs because the denominator, \(x+1\), becomes very large, reducing the fraction \(\frac{1}{x+1}\) to close to zero. The square root of a value approaching zero also tends to zero.

In essence, limits are a crucial tool in understanding a function's end behavior as they provide insights into the function's continuity and overall graph shape.

Square roots in functions

Square roots often introduce complexity when determining the domain of a function. A square root function, like \(f(x) = \sqrt{\frac{1}{x+1}}\), is defined only for non-negative values within the square root.

The expression under the square root symbol, namely \(\frac{1}{x+1}\), must be greater than or equal to zero. It implies that \(x+1\) must be positive, which aside from zero in the denominator, does not introduce further domain restrictions here. However, in other functions, if the numerator itself is complex, it might require additional analysis to understand the root behavior, like the function \(g(x) = \sqrt{\frac{x}{x+1}}\). Here, \(x\) also needs to be non-negative to keep the square root expression valid. This adds the condition \(x \geq 0\) to the domain constraints.

These conditions influence what values \(x\) can take in the real number line, ensuring the square root results in real values, and it highlights the importance of always checking the domain, especially with roots involved.

Rational functions

Rational functions are expressions formed by the ratio of two polynomials. They bring attention to the restrictions caused by the denominator because divisions by zero are undefined in mathematics.

Take the example of \(f(x) = \sqrt{\frac{1}{x+1}}\). The rational part, \(\frac{1}{x+1}\), dictates that \(x\) cannot be -1. This is because \(x+1 = 0\) at \(x = -1\) would cause division by zero, making the function undefined. Similarly, \(g(x) = \sqrt{\frac{x}{x+1}}\) has the same restriction. However, the constraint for \(x\) being non-negative introduces an additional limitation. So, the full domain of \(g(x)\) becomes the intersection of all these conditions, requiring that \(x eq -1\) and \(x \geq 0\).

The nature of these functions requires keeping a close eye on both the numerator and the denominator to ensure valid function outputs. This analysis expands our understanding by highlighting limitations not just caused by the mathematical functions themselves but also by the context they exist within. Rational functions serve to further unravel the complex relationships within function analysis.