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In Problems \(13-16\), find the functions \(f \circ g\) and \(g \circ f\) and give their domains. $$ f(x)=x^{2}-x+5, g(x)=-x+4 $$

Short Answer

Expert verified
\(f \circ g(x) = x^2 - 7x + 17\) and \(g \circ f(x) = -x^2 + x - 1\); both have domain \(\mathbb{R}\).

Step by step solution

01

Understand Composition of Functions

To find the function composition, you need to substitute the entire function of one (say, \( g(x) \)) into every instance of the variable \( x \) in the other function (say, \( f(x) \)). So for \( f \circ g \), replace \( x \) in \( f(x) \) with \( g(x) \), and for \( g \circ f \), replace \( x \) in \( g(x) \) with \( f(x) \).
02

Find \( f \circ g \)

Substitute \( g(x) = -x + 4 \) into \( f(x) = x^2 - x + 5 \):\[f(g(x)) = ((-x + 4)^2) - (-x + 4) + 5.\]Now, expand and simplify:1. Expand \((g(x))^2: ((-x + 4)^2 = x^2 - 8x + 16.\)2. Simplify the function:\[f(g(x)) = x^2 - 8x + 16 + x - 4 + 5 = x^2 - 7x + 17.\]
03

Find \( g \circ f \)

Substitute \( f(x) = x^2 - x + 5 \) into \( g(x) = -x + 4 \):\[g(f(x)) = -(x^2 - x + 5) + 4.\]Expand and simplify:1. Distribute the negative:\[g(f(x)) = -x^2 + x - 5 + 4.\]2. Simplify the function:\[g(f(x)) = -x^2 + x - 1.\]
04

Determine Domains of Composite Functions

Start by identifying the domain of the original functions \( f(x) \) and \( g(x) \). Both functions are polynomials with no restricted domain, meaning their domain is all real numbers (\( \mathbb{R} \)).Since the functions involve only polynomial operations, the domain of both compositions \( f \circ g(x) \) and \( g \circ f(x) \) is also all real numbers (\( \mathbb{R} \)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Functions
When dealing with composite functions, think of them like a recipe where the output of one function becomes the input for another. This process creates a new function that combines the operations of both functions involved.

For example, if we have two functions, say \( f(x) \) and \( g(x) \), the composition of these functions is written as \( f \circ g(x) \). This means you apply \( g(x) \) first, and then put that result into \( f(x) \). This can be represented mathematically as:
  • \( f \circ g(x) = f(g(x)) \)
Similarly, the opposite order, \( g \circ f(x) \), means you first apply \( f(x) \) and then use that output as the input for \( g(x) \):
  • \( g \circ f(x) = g(f(x)) \)
This nifty process allows mathematicians and students to explore how combining different operations can produce varied results.

In our exercise, composite functions were used to combine the polynomial functions \( f(x) = x^2 - x + 5 \) and \( g(x) = -x + 4 \). Each step in forming the composite function involves substituting and carefully simplifying to reach clean, simplified expressions.
Domains of Functions
The domain of a function is a set of all possible input values (or 'x-values') that will not cause the function to malfunction. Put simply, it includes all the numbers you can plug into a function to get a legitimate output.

Generally, different functions have different domains. Some can take all real numbers, whereas others are more restrictive. For example:
  • Polynomial functions, like \( f(x) = x^2 - x + 5 \) and \( g(x) = -x + 4 \), can handle any real number as input. They have no restrictions like square roots or fractions with variables in the denominator that could limit input values.
  • For composite functions like \( f \circ g(x) \) and \( g \circ f(x) \), their domains are dictated by the original functions involved. Since both \( f(x) \) and \( g(x) \) are polynomials with domains of all real numbers \( (\mathbb{R}) \), the domain of their composite functions is also all real numbers.
Understanding domains is key in math, as it ensures that functions are applied correctly and results are valid.
Polynomials
Polynomials are mathematical expressions that come from adding and subtracting powers of a variable, usually denoted as \( x \). These powers are non-negative whole numbers and can include coefficients, which are the numbers before the variables.

A general form of a polynomial is:\[ a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \]where:
  • \( a_n, a_{n-1}, ..., a_0 \) are coefficients.
  • \( x^n, x^{n-1}, ..., x^0 \) are terms in decreasing powers.
Polynomials are very flexible. They can model a wide range of real-world and theoretical situations. For example, quadratic polynomials like \( f(x) = x^2 - x + 5 \) may represent anything from the trajectory of a thrown ball to the economics of profit functions.

What makes polynomials particularly appealing in calculus and algebra is their smoothness and predictability; you can perform operations such as addition, subtraction, and even composition without running into undefined values, making them a delight to work with.

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Most popular questions from this chapter

In Problems \(51-54\), describe in words how the graphs of the given functions differ. [Hint: Factor and cancel.] $$ \begin{array}{ll} f(x)=-\frac{x^{2}-7 x+6}{x-1}, & g(x)=\left\\{\begin{array}{ll} -\frac{x^{2}-7 x+6}{x-1}, & x \neq 1 \\ 8, & x=1 . \end{array}\right. \\ h(x)=\left\\{\begin{array}{ll} -\frac{x^{2}-7 x+6}{x-1}, & x \neq 1 \\ 5, & x=1 \end{array}\right. \\ f(x)=\frac{x^{4}-1}{x^{2}-1}, & g(x)=\left\\{\begin{array}{ll} \frac{x^{4}-1}{x^{2}-1}, & x \neq 1 \\ 0, & x=1 \end{array}\right. \end{array} $$

Explain why the graph of a one-to-one function \(f\) can have at most one \(x\) -intercept.

In Problems \(9-34\), sketch the graph of the given piecewise-defined function. Find any \(x\) - and \(y\) intercepts of the graph. Give any numbers at which the function is discontinuous $$ y=\left\\{\begin{array}{ll} x-1, & x<0 \\ x+1, & x \geq 0 \end{array}\right. $$

In Problems \(51-54\), describe in words how the graphs of the given functions differ. [Hint: Factor and cancel.] $$ f(x)=\frac{x^{2}-9}{x-3}, \quad g(x)=\left\\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x \neq 3 \\ 4, & x=3, \end{array} \quad h(x)=\left\\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & x \neq 3 \\ 6, & x=3 \end{array}\right.\right. $$

The function \(f\) is not one-to-one on the given domain but is one-to-one on the restricted domain (the second interval). Find the inverse of the one-to- one function and give its domain. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes. $$ f(x)=\frac{1}{2} \sqrt{4-x^{2}},[-2,2] ;[0,2] $$

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