Problem 24 - The Calculus with Analytic Geometry

Question

Functions \(f\) and \(g\) are defined. In each exercise define \(f \circ g\), and determine all values of \(x\) for which \(f \circ g\) is continuous. $$ f(x)=\sqrt{x} ; g(x)=x+1 $$

Step-by-Step Solution

Answer

The composite function is \(\sqrt{x+1}\) and is continuous for \(x \geq -1\).

Step 1 - Define Composite Function

The composite function \(f \circ g\) is defined as \(f(g(x))\). This means we will substitute \(g(x)\) into \(f(x)\).

Step 2 - Compute \(f(g(x))\)

Substitute \(g(x) = x + 1\) into \(f(x) = \sqrt{x}\): \(f(g(x)) = \sqrt{g(x)} = \sqrt{x + 1}\). Therefore, \(f \circ g(x) = \sqrt{x + 1}\).

Step 3 - Determine Continuity

For \(f \circ g(x)\) to be continuous, the expression inside the square root must be non-negative. This means \(x + 1 \geq 0\).

Step 4 - Solve Inequality

Solve \(x + 1 \geq 0\) to find the domain where the function is continuous: \(x \geq -1\).

Step 5 - Conclusion

The composite function \(f \circ g\) is continuous for all \(x \geq -1\).

Key Concepts

Definition of Composite Functions

Composite functions combine two functions into one by substituting the output of one function into the input of another. For example, if we have functions \(f(x)\) and \(g(x)\), then their composite, denoted as \(f \circ g\), means \(f(g(x))\). To form this composite function, we take the output of \(g(x)\) and use it as the input to \(f(x)\). This is an essential operation in mathematics because it helps to simplify complex relationships between variables.

Square Root Functions

A square root function is a function that involves the square root of a variable, often written as \( \sqrt{x} \). This function only produces real numbers for inputs that are non-negative, because the square root of a negative number is not a real number. For example, \(f(x) = \sqrt{x}\) defines the square root function where \(x \geq 0\). Since square root functions only output the principal (non-negative) root, it’s important to consider the domain of the input variable to ensure the function is defined.

Determining Continuity

A function is continuous if there are no breaks, gaps, or jumps in its graph. For a composite function like \(f \circ g\), we need to ensure that the resulting expression is defined for all values within its domain. When determining the continuity of \(f \circ g\), check the conditions under which both functions are individually continuous, and ensure that the input to the second function (\(f\)) remains valid for all outputs of the first function (\(g\)). Essentially, the expression inside any square root or denominator must yield valid values.

Function Domain

The domain of a function is the complete set of input values (\(x\)) for which the function is defined. In other words, these are the values of \(x\) that you can plug into the function without making the function undefined. For example, if a function is defined as \(f(x) = \sqrt{x}\), the domain is \(x \geq 0\) because square roots of negative numbers are not real and thus not defined in the set of real numbers. When considering composite functions, you need to check the domain of both the outer function and the output of the inner function.

Inequality Solving

Inequality solving is essential to determine the domain of functions involving expressions like square roots. When you have an inequality like \(x + 1 \geq 0\), you solve it by isolating \(x\). In this case:

• \(x + 1 \geq 0\)
• Subtract 1 from both sides: \(x \geq -1\)

This tells us what values \(x\) can take for the expression inside a square root (or any other restricted operation) to be valid. Solving these inequalities helps define the range of inputs where the function can operate correctly without becoming undefined.