Problem 33
Question
Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=2 x+3, \quad g(x)=4 x-1 $$
Step-by-Step Solution
Verified Answer
\(f \circ g(x) = 8x + 1\), \(g \circ f(x) = 8x + 11\), \(f \circ f(x) = 4x + 9\), \(g \circ g(x) = 16x - 5\). All domains are all real numbers.
1Step 1: Understand Function Composition
Function composition involves applying one function to the results of another. We denote the composition of two functions, \(f\) and \(g\), as \((f \circ g)(x) = f(g(x))\) and \((g \circ f)(x) = g(f(x))\). In these expressions, \(f\) and \(g\) are functions.
2Step 2: Compose \(f \circ g\)
To find \(f \circ g\), substitute \(g(x) = 4x - 1\) into \(f(x) = 2x + 3\). So, \(f(g(x)) = f(4x - 1) = 2(4x - 1) + 3\). Simplifying, we get \(f(g(x)) = 8x - 2 + 3 = 8x + 1\). The function \(f \circ g(x) = 8x + 1\). Since both \(f(x)\) and \(g(x)\) are defined for all real \(x\), the domain of \(f \circ g\) is all real numbers.
3Step 3: Compose \(g \circ f\)
To find \(g \circ f\), substitute \(f(x) = 2x + 3\) into \(g(x) = 4x - 1\). So, \(g(f(x)) = g(2x + 3) = 4(2x + 3) - 1\). Simplifying, we get \(g(f(x)) = 8x + 12 - 1 = 8x + 11\). The function \(g \circ f(x) = 8x + 11\). Since both \(f(x)\) and \(g(x)\) are defined for all real \(x\), the domain of \(g \circ f\) is all real numbers.
4Step 4: Compose \(f \circ f\)
To find \(f \circ f\), substitute \(f(x) = 2x + 3\) back into itself. So, \(f(f(x)) = f(2x + 3) = 2(2x + 3) + 3\). This simplifies to \(f(f(x)) = 4x + 6 + 3 = 4x + 9\). The function \(f \circ f(x) = 4x + 9\). The domain of \(f \circ f\) is all real numbers, as \(f(x)\) is defined for all real \(x\).
5Step 5: Compose \(g \circ g\)
To find \(g \circ g\), substitute \(g(x) = 4x - 1\) back into itself. So, \(g(g(x)) = g(4x - 1) = 4(4x - 1) - 1\). This simplifies to \(g(g(x)) = 16x - 4 - 1 = 16x - 5\). The function \(g \circ g(x) = 16x - 5\). The domain of \(g \circ g\) is all real numbers, as \(g(x)\) is defined for all real \(x\).
Key Concepts
Domain of a FunctionPolynomial FunctionReal Numbers
Domain of a Function
The domain of a function refers to all the possible input values, often denoted as "x," for which the function is defined. For a function like the ones we're dealing with here, the domain often includes all real numbers or a specific set of numbers.
- When considering linear functions, like our example functions, these are defined anywhere on the real number line.
- There are no excluded values, that is, there are no values that would make the function undefined, such as divisions by zero or square roots of negative numbers in these polynomial functions.
- Therefore, when both functions in a composition are defined for all real numbers, so is their composition.
Polynomial Function
A polynomial function is a function that can be expressed in the form of a polynomial. For example, the function \(f(x) = 2x + 3\) is a linear polynomial, specifically a first-degree polynomial.
- Polynomial functions can be classified by their degree, the highest power of the variable present in the polynomial. Linear functions have a degree of one.
- The coefficients, such as 2 and 3 in \(f(x) = 2x + 3\), give us the shape and position of the polynomial on the coordinate plane.
- Polynomials are continuous functions, meaning they have no breaks, jumps, or holes in their graphs.
Real Numbers
The set of real numbers includes all the rational and irrational numbers. It's essentially any value that can correspond to a point on an infinite number line.
- Rational numbers are fractions or decimals with a finite or repeating pattern, like 1/2 or 0.333...
- Irrational numbers can't be written as simple fractions, like \(\pi\) or the square root of 2.
- Real numbers are crucial in understanding domains because they describe every possible input we can consider for these types of polynomial functions.
Other exercises in this chapter
Problem 32
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