Problem 30 - A Graphical Approach to Precalculus with Limits

Question

For each pair of fimetions, (a) find ( \(f+g)(x),(f-g)(x),\) and \((f g)(x) ;\) (b) give the domains of the functions in part (a); (c) find \(\frac{t}{x}\) and give its domain; (d) find \(f \circ g\) and give is domain: and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt{2+4 x^{2}}, g(x)=x$$

Step-by-Step Solution

Answer

(a) (f+g)(x) = √(2+4x²) + x, (f-g)(x) = √(2+4x²) - x, (fg)(x) = x√(2+4x²); Domain: ℝ. (b) (f/g)(x) = √(2+4x²)/x; Domain: ℝ \ {0}. (c) f(g(x)), g(f(x)), Domains: ℝ.

Step 1: Addition and Subtraction of Functions

For (a), we need to find \((f+g)(x)\) and \((f-g)(x)\). Start by writing the expressions for each:\[(f+g)(x) = f(x) + g(x) = \sqrt{2+4x^2} + x\]\[(f-g)(x) = f(x) - g(x) = \sqrt{2+4x^2} - x\]These are the expressions for the addition and subtraction of the functions.

Step 2: Multiplication of Functions

Next, find the expression for \((fg)(x)\):\[(fg)(x) = f(x) \times g(x) = \sqrt{2+4x^2} \cdot x = x \cdot \sqrt{2+4x^2}\]

Step 3: Domain of Sum, Difference, and Product

Since \(f(x) = \sqrt{2 + 4x^2}\), the expression under the square root, \(2 + 4x^2\), must be non-negative. As \(4x^2\) is always non-negative, \(2 + 4x^2 \geq 2\), which holds for all real \(x\). Therefore, the domain of \(f(x)\), \(g(x)\), and their operations in (a) is all real numbers: \((-\infty, \infty)\).

Step 4: Division of Functions

For (b), find \(\frac{f}{g}(x)\):\[\frac{f}{g}(x) = \frac{\sqrt{2+4x^2}}{x}\]The domain of this function requires \(x eq 0\) because division by zero is undefined. Hence, the domain is all real numbers except zero: \((-\infty, 0) \cup (0, \infty)\).

Step 5: Composition \(f \circ g\) and its Domain

For (c), find \(f(g(x))\):\[f(g(x)) = f(x) = \sqrt{2+4x^2}\]Here, \(f(g(x))\) is equivalent to \(f(x)\). The domain, as already determined in Step 3, is all real numbers: \((-\infty, \infty)\).

Step 6: Composition \(g \circ f\) and its Domain

Lastly, for (d), find \(g(f(x))\):\[g(f(x)) = f(x) = \sqrt{2+4x^2}\]Here, \(g(f(x))\) is simply \(\sqrt{2+4x^2}\). The domain remains the same as in Step 3: the entire set of real numbers \((-\infty, \infty)\).

Key Concepts

Domain of a Function

The domain of a function is a crucial concept in mathematics, as it tells us all the possible input values (or 'x' values) that a function can accept without causing any mathematical errors, like division by zero or taking the square root of a negative number.

For the function in our exercise, \(f(x) = \sqrt{2 + 4x^2}\), the expression inside the square root, \(2 + 4x^2\), must be greater than or equal to zero because you cannot take the square root of a negative number in real numbers. As \(4x^2\) is always non-negative, \(2 + 4x^2\) is always greater than or equal to 2. This means the domain of \(f(x)\) is all real numbers.

Additionally, when combining functions through algebraic operations like addition, subtraction, multiplication, and division, we must ensure that each individual operation is defined. For instance, the division \(\frac{f(x)}{g(x)}\) brings in an extra condition for its domain: \(g(x) eq 0\), as we can't divide by zero. Hence, its domain will exclude any value that makes \(g(x) = 0\).

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Composition of Functions

Function composition involves substituting one function into another, effectively creating a new function. This can be visualized as if one function is nested inside the other. In notational terms, if \(f(x)\) and \(g(x)\) are functions, then the composition \(f \circ g\) is defined as \(f(g(x))\).

In our case, when we find \(f \circ g\), we substitute \(g(x) = x\) into \(f(x)\), resulting in \(f(g(x)) = f(x)\ = \sqrt{2 + 4x^2}\). Therefore, the function itself does not change due to the nature of \(g(x)\). The domain for the composed function \(f \circ g\) is the same as the domain for \(f(x)\), which we've already determined as all real numbers.

Similarly, for \(g \circ f\), we substitute \(f(x) = \sqrt{2 + 4x^2}\) into \(g(x)\), leading to \(g(f(x)) = \sqrt{2 + 4x^2}\). This shows that no matter how we nest these functions, the expression remains \(\sqrt{2 + 4x^2}\), retaining the same domain. It is important to evaluate the innermost function's domain first, as this directly affects the feasibility of the operation.

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Algebraic Operations on Functions

Algebraic operations on functions include processes such as addition, subtraction, multiplication, and division. These operations allow us to create new functions from existing ones.

For example, with functions \(f(x)\) and \(g(x)\) provided in the exercise, the addition of the functions is expressed as \((f + g)(x) = f(x) + g(x)\), which in this case becomes \(\sqrt{2 + 4x^2} + x\). Subtraction is similar: \((f - g)(x) = f(x) - g(x) = \sqrt{2 + 4x^2} - x\).

Multiplying functions involves simply multiplying their outputs: \((fg)(x) = f(x) \times g(x) = x \cdot \sqrt{2 + 4x^2}\). For division, it's crucial to remember that the function \(g(x)\) must not be zero, as division by zero is undefined. This is translated into \(\frac{f}{g}(x) = \frac{\sqrt{2 + 4x^2}}{x}\). The domain here excludes \(x = 0\) to prevent division by zero.

These operations are powerful tools in algebra. They allow for the merging of multiple functions into cohesive expressions, which can then be analyzed collectively to determine behaviors, such as defining the domain and identifying any restrictions that might apply.