Problem 2 - Thomas Calculus

Question

In Exercises 1 and \(2,\) find the domains and ranges of \(f, g, f+g,\) and $f \cdot g . $$ f(x)=\sqrt{x+1}, \quad g(x)=\sqrt{x-1} $$

Step-by-Step Solution

Answer

Domain of f+g and f⋅g: [1, ∞). Range of f, g, f+g, and f⋅g: [0, ∞).

Step 1: Finding the Domain of f(x)

To find the domain of \(f(x) = \sqrt{x + 1}\), we need the expression under the square root to be non-negative. Hence, \(x + 1 \geq 0\). Solving this inequality, \(x \geq -1\). Therefore, the domain of \(f(x)\) is \([-1, \infty)\).

Step 2: Finding the Range of f(x)

Since \(f(x) = \sqrt{x + 1}\), and the square root function outputs non-negative values, the smallest value \(f(x)\) can take is 0 when \(x = -1\). As \(x\) becomes larger, \(f(x)\) can take all values up to infinity. Therefore, the range of \(f(x)\) is \([0, \infty)\).

Step 3: Finding the Domain of g(x)

To find the domain of \(g(x) = \sqrt{x - 1}\), we require \(x - 1 \geq 0\). Solving this inequality, \(x \geq 1\). Therefore, the domain of \(g(x)\) is \([1, \infty)\).

Step 4: Finding the Range of g(x)

For \(g(x) = \sqrt{x - 1}\), the square root function outputs non-negative values. The minimum value \(g(x)\) can take is 0 when \(x = 1\). As \(x\) increases, \(g(x)\) can take any value up to infinity. Thus, the range of \(g(x)\) is \([0, \infty)\).

Step 5: Finding the Domain of f+g

The function \((f+g)(x) = \sqrt{x+1} + \sqrt{x-1}\) is defined only where both \(f(x)\) and \(g(x)\) are defined. Thus, the domain of \(f+g\) is the intersection of the domains of \(f(x)\) and \(g(x)\), which is \([1, \infty)\).

Step 6: Finding the Range of f+g

Given that \(f(x)\) and \(g(x)\) both have ranges \([0, \infty)\), the range of \(f+g\) is also \([0, \infty)\), as the sum of two non-negative numbers remains non-negative and can potentially reach any non-negative value.

Step 7: Finding the Domain of f⋅g

The function \((f \cdot g)(x) = \sqrt{x+1} \cdot \sqrt{x-1}\) is defined where both \(f(x)\) and \(g(x)\) are defined. The domain of \(f \cdot g\) is the same as \(f+g\), which is \([1, \infty)\).

Step 8: Finding the Range of f⋅g

The range of \(f \cdot g\) is determined by the product of the outputs of \(f(x)\) and \(g(x)\), both of which are in the interval \([0, \infty)\). The smallest value is 0, reached when either \(\sqrt{x+1}\) or \(\sqrt{x-1}\) is 0. As \(x\) increases, the product approaches infinity, so the range is \([0, \infty)\).

Key Concepts

Square Root Function

The square root function is a fundamental mathematical concept that involves finding a value that, when multiplied by itself, produces the given number. In our exercise, we have two functions defined by square roots: \( f(x) = \sqrt{x+1} \) and \( g(x) = \sqrt{x-1} \). Each function takes on non-negative values only, as the square root of a real number is always non-negative.
For a square root function like \( f(x) = \sqrt{x + 1} \), the expression inside the root, \( x + 1 \), must be zero or positive. This requirement ensures that the function outputs real numbers. Therefore, we solve the inequality \( x + 1 \geq 0 \), giving us the domain of \([-1, \infty)\). The smallest value of \( f(x) \) is 0, occurring at \( x = -1 \), expanding up to infinity as \( x \) grows. Hence, the range is \([0, \infty)\).
Similarly, for \( g(x) = \sqrt{x - 1} \), the domain arises from solving \( x - 1 \geq 0 \), leading to \([1, \infty)\). The range remains the same \([0, \infty)\), since it starts at zero and extends upward based on increased \( x \) values.

Function Operations

Function operations involve performing algebraic procedures such as addition, subtraction, multiplication, or division on functions. In this exercise, we add and multiply two functions, \( f(x) = \sqrt{x+1} \) and \( g(x) = \sqrt{x-1} \).

• **Addition**: The function \((f+g)(x) = \sqrt{x+1} + \sqrt{x-1}\) requires both function domains to be valid simultaneously. Therefore, the domain is the intersection of both function domains: \([1, \infty)\). The range remains \([0, \infty)\) since the sum of two non-negative numbers can vary from zero to potentially any non-negative value.


• **Multiplication**: Similarly, the product function \( (f \cdot g)(x) = \sqrt{x+1} \cdot \sqrt{x-1} \) has a domain constrained by both original functions' domains, thus \([1, \infty)\). The range of the product function is also \([0, \infty)\), as it involves the product of non-negative values.

Performing these operations not only adjusts the expressions but also involves finding common domains where both functions exist validly.

Domain and Range Analysis

Understanding the domain and range of functions is crucial in capturing their limits and behavior. A function's domain consists of all possible input values, while the range consists of all essentially possible output values.
For \( f(x) = \sqrt{x+1} \):

• **Domain**: Solve \( x + 1 \geq 0 \) to find \([-1, \infty)\).
• **Range**: Outputs from zero onwards \([0, \infty)\), as the square root never yields negative results.

For \( g(x) = \sqrt{x-1} \):

• **Domain**: Solve \( x - 1 \geq 0\) finding \([1, \infty)\).
• **Range**: It mirrors \( f(x) \) with \([0, \infty)\).

When functions combine, the domain is where both are simultaneously valid. Thus, for functions like \((f+g)(x)\) and \((f \cdot g)(x)\), the domain transitions to \([1, \infty)\). The ranges remain \([0, \infty)\), accommodating any non-negative result possible from these combinations.