Problem 48 - Calculus Volume 1

Question

For the following exercises, for each pair of functions, find a. \((f \circ g)(x)\) and b. \((g \circ f)(x)\) Simplify the results. Find the domain of each of the results. $$ f(x)=|x+1|, g(x)=x^{2}+x-4 $$

Step-by-Step Solution

Answer

\((f \circ g)(x) = |x^2 + x - 3|\) with domain all real numbers. \((g \circ f)(x) = x^2 + 3x - 2\) with domain all real numbers.

Step 1: Understanding the Composition of Functions

In composition of functions, - \((f \circ g)(x)\) means substitute \(g(x)\) into \(f(x)\), - \((g \circ f)(x)\) means substitute \(f(x)\) into \(g(x)\). Let's handle each of these separately.

Step 2: Finding \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x) = x^2 + x - 4\) into \(f(x) = |x + 1|\): \[ (f \circ g)(x) = f(g(x)) = f(x^2 + x - 4) = |(x^2 + x - 4) + 1| = |x^2 + x - 3| \] Therefore, \((f \circ g)(x) = |x^2 + x - 3|\).

Step 3: Finding the Domain of \((f \circ g)(x)\)

The domain of \((f \circ g)(x) = |x^2 + x - 3|\) is all real numbers because a polynomial function is defined for all real numbers and the absolute value function can handle all real inputs.

Step 4: Finding \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x) = |x + 1|\) into \(g(x) = x^2 + x - 4\): \[ (g \circ f)(x) = g(f(x)) = g(|x + 1|) = (|x + 1|)^2 + |x + 1| - 4\] Simplify \[ (g \circ f)(x) = (x + 1)^2 + (x + 1) - 4 = x^2 + 2x + 1 + x + 1 - 4 = x^2 + 3x - 2\] Therefore, \((g \circ f)(x) = x^2 + 3x - 2\).

Step 5: Finding the Domain of \((g \circ f)(x)\)

The domain of \((g \circ f)(x) = x^2 + 3x - 2\) is all real numbers because both polynomial and absolute value functions are defined for all real numbers.

Key Concepts

Polynomial Function

Polynomial functions are fundamental elements in algebra and calculus. A polynomial function is expressed as a sum of powers of the variable, each with a constant coefficient. The general form of a polynomial function of degree 2 is \[ f(x) = ax^2 + bx + c \] where:

• a, b, and c are constants.
• x is the variable.
• The highest power of x determines the degree of the polynomial.

In the context of the given exercise, the function \(g(x) = x^2 + x - 4\) is a quadratic polynomial because the highest degree of x is 2. Such functions have a U-shaped or inverted U-shaped graph, known as a parabola. Polynomial functions are continuous and defined for all real numbers. This makes them especially versatile for modeling a wide range of real-world phenomena, from simple physical motion to complex financial calculations.

Absolute Value Function

The absolute value function is a piecewise function that outputs the non-negative value of a number. This means it reflects negative values into positives while keeping positive values unchanged. It is represented as:\[ f(x) = |x| \] with a more detailed piecewise form being:

• If \( x \geq 0 \), then \(f(x) = x\)
• If \( x < 0 \), then \(f(x) = -x\)

In practical terms, the absolute value of a number is its distance from zero on the number line. This property is useful in situations where ignoring the direction of a number is necessary, such as in measurements of magnitude. In the exercise, \(f(x) = |x+1|\) takes whatever result comes from \(x + 1\) and returns its absolute value. The graph of an absolute value function has a unique V-shape centered at the origin. This makes it easily recognizable and useful for understanding relationships where both positive and negative alterations to a process might have similar impacts.

Domain of a Function

The domain of a function refers to the complete set of possible values of the independent variable that make the function operate. Essentially, it defines the "input" values. For polynomial functions like \(g(x) = x^2 + x - 4\), the domain is all real numbers (\( -\infty, \infty \)). This is because polynomials are capable of accepting any real number as an input without undefined behavior.
Absolute value functions also have domains that include all real numbers. This is evident in the given function \(f(x) = |x+1|\), which can also accept any real value for \(x\).When functions are composed, their domains are determined by the intersection of the domains of the composing functions. Since both \(f(x)\) and \(g(x)\) have domains encompassing all real numbers, their compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) also have these domains. Thus, each composed function is defined for all real values, ensuring they can be graphed or analyzed without encountering restrictions in real-world scenarios.