Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=-x^{2}+1, \quad g(x)=\sqrt{x} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = -x + 1\), domain: \([0, \infty)\); (b) \((g \circ f)(x) = \sqrt{-x^2 + 1}\), domain: \([-1, 1]\).

Step 1: Identify the Composition Functions

First, recall that the composition \((f \circ g)(x)\) means to substitute \(g(x)\) into \(f(x)\), and \((g \circ f)(x)\) means to substitute \(f(x)\) into \(g(x)\). Start by identifying these expressions and their respective calculations step-by-step.

Step 1: Compute \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x) = \sqrt{x}\) into \(f(x) = -x^2 + 1\). This means \((f \circ g)(x) = f(g(x)) = f(\sqrt{x}) = - (\sqrt{x})^2 + 1 = -x + 1\).

Step 2: Determine the Domain of \((f \circ g)(x)\)

The domain of \((f \circ g)(x)\) is determined by considering the restrictions from both \(g(x)\) and the composed function. \(g(x) = \sqrt{x}\) requires \(x \geq 0\). Since \(f(g(x)) = -x + 1\) is defined for all real \(x\), the domain of \((f \circ g)(x)\) is also \([0, \infty)\).

Step 3: Compute \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x) = -x^2 + 1\) into \(g(x) = \sqrt{x}\). This means \((g \circ f)(x) = g(f(x)) = g(-x^2 + 1) = \sqrt{-x^2 + 1}\).

Step 4: Determine the Domain of \((g \circ f)(x)\)

The domain of \((g \circ f)(x)\) is restricted by the condition that the expression under the square root must be non-negative, i.e., \(-x^2 + 1 \geq 0\). Solving \(-x^2 + 1 \geq 0\) gives \(-x^2 \geq -1\) or \(x^2 \leq 1\), which means \(-1 \leq x \leq 1\). Thus, the domain is \([-1, 1]\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually "x") for which the function is defined. Finding the domain is a critical step, especially when dealing with functions like square roots or fractions. These types of functions have specific restrictions:

- For a square root function, the expression under the root must be equal to or greater than zero. This is because the square root of a negative number isn't defined in real numbers.
- For a fraction, the denominator cannot be zero, as division by zero is undefined.

In the given exercise, when finding the domain for the composite functions \((f \circ g)(x)\) and \((g \circ f)(x)\), we consider the individual domains of \(f(x)\) and \(g(x)\), and any new restrictions that arise in the composition.

For \(f \circ g\), since \(g(x) = \sqrt{x}\), \(x\) needs to be \(\geq 0\). The function \(f(x) = -x + 1\) has no further restrictions, so the domain of \((f \circ g)(x)\) is \([0, \infty)\).

For \(g \circ f\), you need \(-x^2 + 1\) to be non-negative, which reduces the domain to \([-1, 1]\), since the square root requires non-negative input.

Composite Function

A composite function is formed when one function is applied to the results of another function. It's denoted as \((f \circ g)(x)\), meaning you take \(g(x)\) and plug it into \(f(x)\). To compute a composite function, simply substitute one function into the other.

For instance, the exercise asked us to find \((f \circ g)(x)\) and \((g \circ f)(x)\).

- For \(f \circ g\), this means plugging \(g(x) = \sqrt{x}\) into \(f(x) = -x^2 + 1\), resulting in \((f \circ g)(x) = -x + 1\).
- For \(g \circ f\), you replace \(f(x)\) with \(-x^2 + 1\) within \(g(x) = \sqrt{x}\), giving \((g \circ f)(x) = \sqrt{-x^2 + 1}\).

When computing, ensure each substitution respects the domain of both functions involved, as this affects the domain of the composite function.

Square Root Function

The square root function, denoted as \(\sqrt{x}\), calculates the non-negative root of \(x\). One of its defining characteristics is the domain restriction: the expression within the square root must be non-negative, meaning \(x \geq 0\). This is because the square root of a negative number is not defined in the realm of real numbers.

In the exercise, the square root function \(g(x) = \sqrt{x}\) was used in two compositions:

• \( \(f \circ g\)(x) = -x + 1\) had its domain dictated by \(g(x)\), limiting \(x \geq 0\).
• \( \(g \circ f\)(x) = \sqrt{-x^2 + 1}\) was constructed such that \(-x^2 + 1\) maintained a non-negative outcome (\(-1 \leq x \leq 1\)).

Understanding the implications of applying a square root function helps you determine proper compositions and their valid domains.