Problem 67
Question
Let \(f(x)=3 x-2\) and \(g(x)=2 x^{2}+1 .\) Find each function and give its domain. $$ f / g $$
Step-by-Step Solution
Verified Answer
The function \( f/g = \frac{3x - 2}{2x^2 + 1} \) is defined for all real numbers.
1Step 1: Understanding the Functions
We have two functions: \( f(x) = 3x - 2 \) and \( g(x) = 2x^2 + 1 \). We will first understand what each function represents individually. \( f(x) \) is a linear function, and \( g(x) \) is a quadratic function.
2Step 2: Identifying the Domains
The domain of a function is the set of all possible input values (\(x\)) it can accept. Since both \( f(x) \) and \( g(x) \) are polynomials, their domain is all real numbers, \( \mathbb{R} \).
3Step 3: Forming the Function \( f/g \)
The expression \( f/g \) is defined as \( \frac{f(x)}{g(x)} \), which gives \( \frac{3x - 2}{2x^2 + 1} \). This expression represents the division of the two functions.
4Step 4: Determining the Domain of \( f/g \)
The domain of \( f/g \) is influenced by the denominator \( g(x) \). Since a rational function is undefined where its denominator is zero, we solve \( 2x^2 + 1 = 0 \). However, \( 2x^2 + 1 > 0 \) for all real \( x \), ensuring the function is never undefined. Thus, the domain remains \( \mathbb{R} \).
5Step 5: Conclusion
Thus, the domain of the function \( f/g \), which is \( \frac{3x - 2}{2x^2 + 1} \), is all real numbers. The function \( f/g \) is a rational function of a linear and a quadratic polynomial.
Key Concepts
Function DomainLinear FunctionsQuadratic Functions
Function Domain
The domain of a function is essentially the complete set of possible values that you can input into the function. It tells you which values of x are acceptable for calculations.
For most polynomial functions, such as linear and quadratic functions, the domain is quite straightforward.
For most polynomial functions, such as linear and quadratic functions, the domain is quite straightforward.
- For the linear function \( f(x) = 3x - 2 \), the domain covers all real numbers \( \mathbb{R} \). This is because there are no restrictions on what x-value you can use for calculating \( 3x - 2 \).
- In the case of the quadratic function \( g(x) = 2x^2 + 1 \), the same principle applies. All real numbers \( \mathbb{R} \) are suitable for this input as well, as a quadratic expression does not have restrictions on x.
Linear Functions
Linear functions are among the simplest types of functions to deal with. Their general form is \( f(x) = mx + b \), where \( m \) represents the slope and \( b \) is the y-intercept.
- The function \( f(x) = 3x - 2 \) is a perfect example of a linear function. Here, "3" is the slope, which means for every unit increase in x, \( f(x) \) increases by 3. "-2" is the y-intercept, which tells us that the line crosses the y-axis at \( -2 \).
- Linear functions graph as straight lines. They continue indefinitely in both directions of the x and y axes. As such, their domain is always all real numbers \( \mathbb{R} \).
Quadratic Functions
Quadratic functions are a step up in complexity from linear functions. They take the general form of \( g(x) = ax^2 + bx + c \).
- The function \( g(x) = 2x^2 + 1 \) is a typical quadratic function where "2" is the coefficient of \( x^2 \), "0" can be assumed for the linear term, and "1" is the constant term.
- Quadratic functions graph as parabolas, which are U-shaped curves. If the coefficient \( a \) is positive, the parabola opens upwards, like this case. If it’s negative, it opens downwards.
- The domain for any quadratic function is all real numbers \( \mathbb{R} \) because you can substitute any real value of x into the quadratic equation without facing undefined operations.
Other exercises in this chapter
Problem 67
Solve each equation. Express all answers to four decimal places. See Example 5. $$ \ln x=1.001 $$
View solution Problem 67
Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log (x+90)+\log x=3 $$
View solution Problem 67
In the definition of the exponential function, \(b\) could not be negative. Why?
View solution Problem 67
Solve for \(x\). See Example 3 . $$ \log _{100} x=\frac{3}{2} $$
View solution