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Chapter 4: Problem 43

Find a linear function \(f(x)=m x+b\) such that \(m\) is positive and \((f \circ f)(x)=9 x+4\)

Short Answer

Expert verified
The function is \(f(x) = 3x + 1\).

Step by step solution

01

Understand the Problem

We need to find a linear function \( f(x) = mx + b \) such that the composition \((f \circ f)(x) = f(f(x))\) equals \(9x + 4\). This means applying the function \(f(x)\) twice should result in a new function that has the form \(9x + 4\).
02

Set Up the Composition

To find \((f \circ f)(x)\), substitute \(f(x)\) into itself: \(f(f(x)) = f(mx + b) = m(mx + b) + b = m^2x + mb + b \). This is the expression we want equal to \(9x + 4\).
03

Compare Coefficients

From step 2, we have \(m^2x + mb + b = 9x + 4\). By comparing coefficients, we equate the coefficients of \(x\) and the constant terms: \(m^2 = 9\) and \(mb + b = 4\).
04

Solve for \(m\)

The equation \(m^2 = 9\) gives us \(m = 3\) or \(m = -3\). However, since \(m\) is positive, we choose \(m = 3\).
05

Solve for \(b\)

Substitute \(m = 3\) into the equation \(mb + b = 4\), which becomes \(3b + b = 4\). Simplifying gives \(4b = 4\), so \(b = 1\).
06

Write the Function

Having determined \(m = 3\) and \(b = 1\), the linear function is \(f(x) = 3x + 1\).

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Key Concepts

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

Function Composition
Function composition involves applying one function to the results of another. If we have two functions, say, \(g(x)\) and \(h(x)\), then the composition \((g \circ h)(x)\) translates to \(g(h(x))\). It鈥檚 like a chain process where the output of function \(h(x)\) becomes the input of function \(g(x)\). In many mathematical problems, especially in calculus and precalculus, function composition helps create complex relationships between different equations.
  • Example: If \(g(x) = 2x\) and \(h(x) = x + 3\), then \((g \circ h)(x) = g(h(x)) = g(x + 3) = 2(x + 3) = 2x + 6\).
  • Function composition is not always commutative; that is, \((g \circ h)(x)\) can be different from \((h \circ g)(x)\).

In our exercise, the function composition \((f \circ f)(x)\) translates to \(f(f(x))\). This means plugging \(f(x) = mx + b\) into itself resulting in \(m(mx + b) + b\). This setup is crucial for solving the given problem, which requires us to express this form in terms of \(9x + 4\).
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true. In algebra, especially when dealing with linear equations, we often end up with systems of equations or need to figure out specific values. In the problem, we solve equations by comparing coefficients and isolating variables.
  • The equation \(m^2 x + mb + b = 9x + 4\) presents both variables \(m\) and \(b\) to be solved.
  • This required setting \(m^2 = 9\) and \(mb + b = 4\).
  • These equations help us solve for \(m\) and \(b\) by cleverly breaking down similar terms, showcasing the versatility of linear equation strategies.

By isolating equations and examining constant and linear terms separately, we ensure stepwise simplification. First, we determined that \(m = 3\) since \(m\) must be positive, then using \(m = 3\), we substituted back to solve for \(b = 1\). These steps form the core of solving linear equations algebraically.
Precalculus Concepts
Precalculus serves as a transition between algebra and calculus, introducing essential concepts that build a solid foundation. Our problem utilizes key precalculus concepts like linear functions, function composition, and the manipulation of algebraic expressions. Understanding these principles is pivotal in solving more intricate problems later on.
  • Linear Functions: In the given task, a linear function is expressed in the form \(f(x) = mx + b\). Recognizing this format, allows us to equate and solve based on given conditions.
  • Function Composition: Used frequently in calculus, it emphasizes working through nested functions, a skill initially built in precalculus.
  • Algebraic Manipulations: The exercise also highlights handling multiple variables and equations, crucial for any calculus-based operation.

Precalculus equips students with the tools to navigate these areas confidently, bridging simpler algebra concepts with more advanced calculus applications. By practicing these ideas through exercises like ours, students reinforce their understanding and are better prepared for mathematical challenges in calculus and beyond.

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Most popular questions from this chapter

Suppose that \(a\) and \(b\) are positive numbers whose sum is 1. (a) Find the maximum possible value of the product ab. (b) Prove that \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right) \geq 9\)

Let \(g(x)=x^{2}+\) bx. Are there any values for \(b\) for which the minimum value of this function is \(-1 ?\) If so, what are they? If not, explain why.

(a) Determine the \(x\) - and \(y\) -intercepts and the excluded regions for the graph of the given function. Specify your results using a sketch similar to Figure \(16(a)\) (b) Describe the behavior of the function at each \(x\) -intercept that corresponds to a repeated factor. Specify your results using a sketch similar to the left-hand portion of Figure \(20 .\) (c) Graph each function. $$y=(x+1)^{2}(x-1)(x-3)$$

Let \(f(x)=\left(x^{5}+1\right) / x^{2}\) (a) Graph the function \(f\) using a viewing rectangle that extends from -4 to 4 in the \(x\) -direction and from -8 to 8 in the \(y\) -direction. (b) Add the graph of the curve \(y=x^{3}\) to your picture in part (a). Note that as \(|x|\) increases (that is, as \(x\) moves away from the origin), the graph of \(f\) looks more and more like the curve \(y=x^{3} .\) For additional perspective, first change the viewing rectangle so that \(y\) extends from -20 to \(20 .\) (Retain the \(x\) -settings for the moment.) Describe what you see. Next, adjust the viewing rectangle so that \(x\) extends from -10 to 10 and \(y\) extends from -100 to \(100 .\) Summarize your observations. (c) In the text we said that a line is an asymptote for a curve if the distance between the line and the curve approaches zero as we move further and further out along the curve. The work in part (b) illustrates that a curve can behave like an asymptote for another curve. In particular, part (b) illustrates that the distance between the curve \(y=x^{3}\) and the graph of the given function \(f\) approaches zero as we move further and further out along the graph of \(f .\) That is, the curve \(y=x^{3}\) is an "asymptote" for the graph of the given function \(f\). Complete the following two tables for a numerical perspective on this. In the tables, \(d\) denotes the vertical distance between the curve \(y=x^{3}\) and the graph of \(f:\) $$ d=\left|\frac{x^{5}+1}{x^{2}}-x^{3}\right| $$ $$\begin{array}{llllll} \hline x & 5 & 10 & 50 & 100 & 500 \\ \hline d & & & & \\ \hline & & & & \\ \hline x & -5 & -10 & -50 & -100 & -500 \\ \hline d & & & & \\ \hline \end{array}$$ (d) Parts (b) and (c) have provided both a graphical and a numerical perspective. For an algebraic perspective that ties together the previous results, verify the following identity, and then use it to explain why the results in parts (b) and (c) were inevitable: $$ \frac{x^{5}+1}{x^{2}}=x^{3}+\frac{1}{x^{2}} $$

Let \(A B C\) be a triangle with vertices \(A(-3,0), B(0,6)\) \(C(3,0) .\) Let \(P\) be a point on the line segment that joins \(B\) to the origin. Find the position of \(P\) that minimizes the sum of the distances between \(P\) and the vertices.
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