Problem 41
Question
For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). \(h(x)=\sqrt{\frac{2 x-1}{3 x+4}}\)
Step-by-Step Solution
Verified Answer
\(f(x) = \sqrt{x}\) and \(g(x) = \frac{2x-1}{3x+4}\).
1Step 1: Identify Outer and Inner Functions
To express the function as a composition, we need to identify the outer function and the inner function. For the function \(h(x) = \sqrt{\frac{2x - 1}{3x + 4}}\), the outer function is a square root function, \(f(u) = \sqrt{u}\).
2Step 2: Define the Inner Function
The inner function \(g(x)\) is the expression inside the square root. So we let \(g(x) = \frac{2x - 1}{3x + 4}\). This transforms the exam question function into a composition \(h(x) = f(g(x))\).
3Step 3: Express in Composite Function Form
We've determined that \(f(g(x)) = \sqrt{g(x)}\) where \(g(x) = \frac{2x - 1}{3x + 4}\). Thus, \(f(x)\) is \(\sqrt{x}\), and \(g(x)\) is \(\frac{2x - 1}{3x + 4}\).
Key Concepts
Understanding Algebraic ExpressionsBreaking Down Composite FunctionsExploring Square Root Functions
Understanding Algebraic Expressions
Algebraic expressions form the foundation of many mathematical concepts. They consist of numbers, variables, and operations like addition, subtraction, multiplication, and division. Understanding these expressions is crucial when dealing with functions like the one in this exercise.
Let's break down the given expression inside the square root to better understand it: \(\frac{2x - 1}{3x + 4}\). This is a rational algebraic expression, meaning it is the ratio of two polynomials.
The numerator \(2x - 1\) and the denominator \(3x + 4\) each represent a linear polynomial. When working with algebraic expressions in the context of functions, it's essential to know several things:
Let's break down the given expression inside the square root to better understand it: \(\frac{2x - 1}{3x + 4}\). This is a rational algebraic expression, meaning it is the ratio of two polynomials.
The numerator \(2x - 1\) and the denominator \(3x + 4\) each represent a linear polynomial. When working with algebraic expressions in the context of functions, it's essential to know several things:
- **Simplification:** Can the expression be simplified? In our case, the expression is already in its simplest form.
- **Evaluation:** What happens if we substitute a specific value for \(x?\) Evaluate \(\frac{2x - 1}{3x + 4}\) carefully.
- **Domain:** What are the allowable values of \(x?\) Avoid values that make the denominator zero, like \(-\frac{4}{3}\).
Breaking Down Composite Functions
Composite functions are made from combining two or more functions into one. They are written as \(f(g(x))\), meaning the output of one function \(g(x)\) becomes the input to another function \(f(x)\).
In the exercise, we identified the composite function \(h(x) = \sqrt{\frac{2x - 1}{3x + 4}}\) as \(h(x) = f(g(x))\). Here, our outer function \(f(u)\) is a square root function, while our inner function \(g(x)\) is a rational expression.
It's essential to understand the step-by-step process of forming composite functions:
In the exercise, we identified the composite function \(h(x) = \sqrt{\frac{2x - 1}{3x + 4}}\) as \(h(x) = f(g(x))\). Here, our outer function \(f(u)\) is a square root function, while our inner function \(g(x)\) is a rational expression.
It's essential to understand the step-by-step process of forming composite functions:
- **Identify components:** Break down the function into its constituent parts.
- **Substitution:** Implement substitution by replacing \(g(x)\) into \(f(x)\) as \(f(g(x))\).
- **Evaluate:** Evaluate the composite function for specific \(x\) values, taking note of any restrictions caused by the inner function.
Exploring Square Root Functions
Square root functions involve the square root operation, often appearing in various mathematical and real-world scenarios. They have the general form \(f(x) = \sqrt{x}\), where \(x\) must be non-negative to have real number outputs.
In this exercise, the square root function \(f(u) = \sqrt{u}\) acts as the outer function in the composite setup. Square root functions have unique properties:
In this exercise, the square root function \(f(u) = \sqrt{u}\) acts as the outer function in the composite setup. Square root functions have unique properties:
- **Domain and Range:** The domain consists of non-negative values for \(x\). The range consists of non-negative real numbers.
- **Behavior:** \(f(x)\) increases slowly for large \(x\). The curve becomes less steep as \(x\) grows.
- **Graphical representation:** It is half of a parabola lying sideways.
Other exercises in this chapter
Problem 40
For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increa
View solution Problem 41
A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using \(x\) as the diameter of the bearing, write this stat
View solution Problem 41
For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. \(f(x)=\left\\{\begin{array}{cl}3 & \text { if } x
View solution Problem 42
For the following exercises, find the inverse function. Then, graph the function and its inverse. \(f(x)=\frac{3}{x-2}\)
View solution