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Chapter 3: Problem 42

Write the given function as the composite of two functions, neither of which is the identity function, as in Examples 6 and 7 . (There may be more than one way to do this.) $$f(x)=\frac{1}{3 x^{2}+5 x-7}$$

Short Answer

Expert verified
Question: Express the function $$f(x)=\frac{1}{3 x^{2}+5 x-7}$$ as a composite of two functions g(x) and h(x) such that no identity functions should be present in the composition. Answer: One possible way is to represent it as $$f(x)=g(h(x))$$ where $$g(x)=\frac{1}{x}$$ and $$h(x)=3x^2+5x-7$$. Another representation is $$f(x)=h(g^{-1}(x))$$ with $$g(x)=\frac{x}{3}$$, $$g^{-1}(x)=3x$$, and $$h(x)=\frac{1}{x^2+5x-7}$$.

Step by step solution

01

Way 1: Break f(x) into quotient and denominator functions

Let's express the given function as a composite of two functions - one as the numerator, and another as the denominator. We can define the functions as: $$g(x) = \frac{1}{x},$$ $$h(x) = 3x^2 + 5x - 7.$$ Now check if $$f(x)= g(h(x))$$: $$f(x) = g(h(x)) = g(3x^2 + 5x - 7) = \frac{1}{3x^2 + 5x - 7}.$$ This representation satisfies the given conditions.
02

Way 2: Break f(x) into a function and its inverse

Another way to represent the given function as a composite of two functions can be using a function and its inverse. Let's first define the function $$g(x) = \frac{x}{3},$$ and its inverse is $$g^{-1}(x) = 3x.$$ Now, define the function h(x) as: $$h(x) = \frac{1}{x^2 + 5x - 7}.$$ Now check if $$f(x)= h(g^{-1}(x))$$: $$f(x) = h(g^{-1}(x)) = h(3x) = \frac{1}{(3x)^2 + 5(3x) - 7} = \frac{1}{3x^2 + 5x - 7}.$$ This representation also satisfies the conditions given.

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Key Concepts

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

Composite Function
Composite functions are formed when one function is applied to the result of another function. It鈥檚 like layers in a cake: you apply the first function, and then the second one uses the result of the first. For our exercise, we have two primary ways of forming composite functions using the given function.
  • First Method: We took the internal function, denoted as \(h(x)\), which was \(3x^2 + 5x - 7\), and used it inside function \(g(x)\), which was \(\frac{1}{x}\). This gave us \(f(x) = g(h(x)) = \frac{1}{3x^2 + 5x - 7}\).
  • Second Method: We used function \(g(x) = \frac{x}{3}\) and its inverse to create the composite \(f(x) = h(g^{-1}(x))\).
These methods showed two valid ways to express the original function \(\frac{1}{3 x^{2}+5x-7}\) as a composite of two functions. Neither approach used the identity function (a function that returns the input as output), which was necessary to meet the problem's conditions.
Function Inverse
An inverse function reverses the operation of the original function. If you apply a function to an input and then its inverse to the result, you end up with the original input. In mathematical terms, for a function \(g(x)\) and its inverse \(g^{-1}(x)\), the property \(g(g^{-1}(x)) = x\) holds true.
  • For example, in the exercise, we defined \(g(x) = \frac{x}{3}\). Its inverse, \(g^{-1}(x) = 3x\), reverses the operation of dividing by 3, effectively multiplying by 3 to return the original value.
  • To create the composite function using inverses, we composed \(h(g^{-1}(x))\) in the expression \(f(x) = \frac{1}{(3x)^2 + 5 \times 3x - 7}\).
Understanding inverses is crucial in functions, since they help reverse processes and solve equations.
Algebraic Expression
Algebraic expressions are foundational elements of algebra, consisting of variables, numbers, and operations. In the exercise, we used the expression \(3x^2 + 5x - 7\) as part of our composite functions.
  • These expressions often form the "inside" function (like \(h(x)\) in our problem), which is then manipulated or transformed by an outer function.
  • Understanding how to manipulate algebraic expressions, such as factoring or expanding, is essential in forming and simplifying complex expressions or functions.
Harnessing algebraic expressions is key when working with functions, as they allow us to model and solve real-world problems efficiently.

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Most popular questions from this chapter

Write the given function as the composite of two functions, neither of which is the identity function, as in Examples 6 and 7 . (There may be more than one way to do this.) $$f(x)=\sqrt[3]{x^{2}+2}$$

Determine the domain of the function according to the usual convention. $$h(x)=\frac{\sqrt{x-1}}{x^{2}-1}$$

Find the radius \(r\) and height \(h\) of a cylindrical can with a surface area of 60 square inches and the largest possible volume, as follows. (a) Write an equation for the volume \(V\) of the can in terms of \(r\) and \(h\). (b) Write an equation in \(r\) and \(h\) that expresses the fact that the surface area of the can is \(60 .\) [ Hint: Think of cutting the top and bottom off the can; then cut the side of the can lengthwise and roll it out flat; it's now a rectangle. The surface area is the area of the top and bottom plus the area of this rectangle. The length of the rectangle is the same as the circumference of the original can (why?).] (c) Write an equation that expresses \(V\) as a function of \(r\) [Hint: Solve the equation in part (b) for \(h\), and substitute the result in the equation of part (a).] (d) Graph the function in part (c), and find the value of \(r\) that produces the largest possible value of \(V\). What is \(h\) in this case?

Find the average rate of change of the function f over the given interval. $$f(x)=-\sqrt{x^{4}-x^{3}+2 x^{2}-x+4} \text { from } x=0 \text { to } x=3$$

Fill the blanks in the given table. In each case the values of the functions \(f\) and \(g\) are given by these tables: $$\begin{array}{|c|c|} \hline x & f(x) \\ \hline 1 & 3 \\ \hline 2 & 5 \\\ \hline 3 & 1 \\ \hline 4 & 2 \\ \hline 5 & 3 \\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 5 \\ \hline 2 & 4 \\\ \hline 3 & 4 \\ \hline 4 & 3 \\ \hline 5 & 2 \\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline t & (f \circ g)(t) \\ \hline 1 & \\ \hline 2 & 2 \\ \hline 3 & \\ \hline 4 & \\ \hline 5 & \\ \hline \end{array}$$
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