Problem 19 - Precalculus

Question

In Exercises \(11-20\), use the pair of functions \(f\) and \(g\) to find the domain of the indicated function then find and simplify an expression for it. $$ f(x)=x \text { and } g(x)=\sqrt{x+1} $$

Step-by-Step Solution

Answer

The domain of \((f \circ g)(x)\) is \(x \geq -1\), with the simplified expression \(\sqrt{x+1}\).

Step 1: Identify the Indicated Function

To determine which function we need to find the domain for, determine the indicated operation on functions. Since no specific combination is mentioned, we assume the function operation is the most common one: composition. Therefore, we find the domain for the composition function \((f \circ g)(x) = f(g(x))\).

Step 2: Find Formula for the Composition Function

The composition \((f \circ g)(x)\) means we substitute \(g(x)\) into \(f\). We know \(f(x) = x\), so \((f \circ g)(x) = g(x)\), which simplifies to \(\sqrt{x+1}\).

Step 3: Determine the Domain of the Composition Function

The function \(g(x) = \sqrt{x+1}\) is only defined when the expression under the square root is non-negative. Therefore, set up the inequality: \(x + 1 \geq 0\). Solve for \(x\): \(x \geq -1\). The domain of the composition function is all real numbers \(x\) such that \(x \geq -1\).

Step 4: Simplify the Expression

From Step 2, the simplified form of the indicated function composition \((f \circ g)(x)\) is \(\sqrt{x + 1}\), which is already as simplified as possible.

Key Concepts

Function Composition

Creating a new function by combining two existing functions is called function composition. It involves feeding the output of one function into the input of another. In our problem, we have two functions: \( f(x) = x \) and \( g(x) = \sqrt{x+1} \). To compose these functions, we perform the operation \((f \circ g)(x) = f(g(x))\). This simply means we take \( g(x) \) and use it as the input for \( f(x) \). In this case, since \( f(x) = x \), the composition \((f \circ g)(x)\) results in \( \sqrt{x+1} \). This step is crucial for understanding how functions can be transformed and manipulated together to form new expressions.

Square Root Function

The square root function \( g(x) = \sqrt{x+1} \) is a classic example of a function where we need to be mindful of its domain. Square root functions are defined for non-negative numbers, which means the expression under the square root must be greater than or equal to zero. This is because the square root of a negative number does not yield a real number.

• Example: \( g(x) = \sqrt{x+1} \) is defined when \( x+1 \geq 0 \).
• Reason: We cannot take the square root of a negative result in the context of real numbers.

Understanding how to find the domain of a square root function can help in accurately determining the values of \( x \) for which the function is defined.

Domain Restriction

Domain restriction is the process of narrowing down the range of possible input values for a function, based on its mathematical properties. For the composed function \( \sqrt{x+1} \), we determine the domain by ensuring that \( x+1 \geq 0 \). To find this restriction:

• Set up the inequality: \( x+1 \geq 0 \).
• Simplify to find \( x \geq -1 \).

The domain of this function, therefore, includes all real numbers \( x \) such that \( x \geq -1 \). This is crucial because it defines where the function is valid and can produce real number outputs.

Function Simplification

Function simplification is about reducing an expression to its most basic form without changing its value. For our composed function, \( (f \circ g)(x) = \sqrt{x+1} \), the expression is already in its simplest form. Simplifying functions involves removing unnecessary complexity and can sometimes involve factoring, canceling terms, or reformatting.

• Simplified expressions are easier to evaluate and interpret.
• No further simplification is needed when an expression reaches its simplest functional form, as in the case here.

Simplified forms aid in understanding the essence of what a function shows and in solving related problems more efficiently.