/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Let \(f(x)=x^{2}+3 x\) and \(g(x... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)=x^{2}+3 x\) and \(g(x)=2 x-1 .\) Perform the composition or operation indicated. $$(f+g)(3)$$

Short Answer

Expert verified
\((f+g)(3) = 23\).

Step by step solution

01

Understand the Operation

The problem requires us to find the value of \((f+g)(3)\), which means we need to first determine the function \((f+g)(x)\) and then evaluate it at \(x=3\).
02

Determine \((f+g)(x)\)

The function \((f+g)(x)\) is found by adding the functions \(f(x)\) and \(g(x)\).So, \((f+g)(x) = f(x) + g(x) = x^2 + 3x + (2x - 1)\).Combine the like terms:\((f+g)(x) = x^2 + 5x - 1\).
03

Evaluate \((f+g)(3)\)

Now, substitute \(x = 3\) into \((f+g)(x) = x^2 + 5x - 1\):\((f+g)(3) = 3^2 + 5(3) - 1\).
04

Compute the Result

Calculate the expression:\(3^2 = 9\),\(5(3) = 15\),Add these results: \(9 + 15 = 24\),Subtract \(1\):\(24 - 1 = 23\).

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Key Concepts

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

Algebraic Functions
Algebraic functions are expressions made up of a combination of numbers, variables, and operations such as addition, subtraction, multiplication, division, and taking roots. These functions can be simple, like linear functions, or more complex, like polynomial functions.

In this exercise, two algebraic functions are given, specifically a polynomial function, \( f(x) = x^2 + 3x \), and a linear function, \( g(x) = 2x - 1 \). Together, they form what is called a composite function when they are combined in operations, such as addition.

Think of algebraic functions like a recipe: they tell you how to combine different ingredients (numbers and variables) to create a desired result. Understanding how to manipulate these functions is crucial in algebra, as it allows us to solve equations and work with complex mathematical models efficiently.
Function Evaluation
The concept of function evaluation involves finding the output of a function given a specific input. To evaluate a function, you substitute the given value into the function and simplify the expression to find the result. In the step-by-step solution of the original exercise, you begin by first combining the functions \( f(x) \) and \( g(x) \) to form \( (f+g)(x) \).

Once you have the combined function, you'll plug in the value of 3 for \( x \) to find \( (f+g)(3) \). This process highlights the power of function evaluation: it converts abstract expressions into concrete numbers.

Function evaluation is akin to following a set of instructions where the input, like \( x = 3 \), guides you through the calculation process to reach an output, offering clear and tangible results. This concept is fundamental in algebra and many other mathematics disciplines.
Polynomial Functions
Polynomial functions are a specific type of algebraic function, consisting of terms that are powers of a variable, typically written as \( ax^n + bx^{n-1} + \, ... \, + c \), where \( n \) is a non-negative integer, and \( a, b, \) and \( c \) are constants.

In our example, \( f(x) = x^2 + 3x \) is a polynomial function because it is composed of terms like \( x^2 \) and \( 3x \). This means that polynomial functions consist of sums of monomials, where each term acts as a building block added to create a complete function.

Understanding polynomials is crucial because they serve as the foundation for more complex mathematical structures and are frequently encountered in algebra. They can be easily manipulated through operations like addition, subtraction, and even differentiation and integration, which is why mastering them is essential for anyone studying advanced mathematics.

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Most popular questions from this chapter

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The graphing calculator screen on the left shows three functions: \(\mathrm{Y}_{1}, \mathrm{Y}_{2},\) and \(\mathrm{Y}_{3} .\) The last of these, \(\mathrm{Y}_{3}\), is defined as \(\mathrm{Y}_{1} \circ \mathrm{Y}_{2},\) indicated by the notation \(\mathrm{Y}_{3}=\mathrm{Y}_{1}\left(\mathrm{Y}_{2}\right) .\) The table on the right shows selected values of \(\mathbf{X},\) along with the calculated values of \(\mathbf{Y}_{3} .\) Predict the display for \(\mathbf{Y}_{3}\) for the given value of \(\mathbf{X}\). $$\mathbf{X}=-1$$

During the early years of the AIDS epidemic, cases and cumulative deaths reported for selected years \(x\) could be modeled by quadratic functions. For \(1982-\) 1994 , the numbers of AIDS cases are modeled by $$f(x)=3200(x-1982)^{2}+1586$$ and the numbers of deaths are modeled by $$g(x)=1900(x-1982)^{2}+619$$ $$\begin{array}{|l|c|c|}\hline \text { Year } & \text { Cases } & \text { Deaths } \\\\\hline 1982 & 1,586 & 619 \\ 1984 & 10,927 & 5,605 \\\1986 & 41,910 & 24,593 \\\1988 & 106,304 & 61,911 \\\1990 & 196,576 & 120,811 \\ 1992 & 329,205 & 196,283 \\\1994 & 441,528 & 270,533\end{array}$$ (a) Graph \(h(x)=\frac{g(x)}{f(x)}\) in the window \([1982,1994]\) by \([0,1] .\) Interpret the graph. (b) Compute the ratio \(\frac{\text { deaths }}{\text { cases }}\) for each year. Compare the results with those from part (a).

Sketch a graph showing the distance a person is from home after \(x\) hours if he or she drives on a straight road at 40 mph to a park 20 miles away, remains at the park for 2 hours, and then returns home at a speed of \(20 \mathrm{mph}\).

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