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Chapter 2: Problem 6

Given \(g(x)=|2 x|-x^{2}+3,\) find \(g(-x)\).

Short Answer

Expert verified
The function remains the same: \( g(-x) = g(x) = 2|x| - x^{2} + 3 \).

Step by step solution

01

Substitute \( -x \) for \( x \) in the original function

Given the function \( g(x) = |2x| - x^{2} + 3 \), to find \( g(-x) \), we replace \( x \) with \( -x \) in the function. So, \( g(-x) = |2(-x)| - (-x)^{2} + 3 \).
02

Simplify the absolute value term

The absolute value of \( 2(-x) \) simplifies as follows: \( |2(-x)| = |-2x| = 2|x| = 2x \) (since \( |x| \) is always positive). Thus, \( g(-x) = 2|x| - (-x)^{2} + 3 \).
03

Simplify the square term

Next, simplify the square term \( (-x)^{2} \), which is the same as \( x^{2} \). So the function becomes \( g(-x) = 2|x| - x^{2} + 3 \).
04

Combine the results

Putting all simplified terms together, the function remains \( g(-x) = 2|x| - x^{2} + 3 \). Notice this is the same as the original function.

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Key Concepts

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

Absolute Value
Understanding absolute value is crucial in college algebra. The absolute value of a number is its distance from zero on the number line, regardless of direction. For any real number, the absolute value of that number is always non-negative.

For example, the absolute value of both -3 and 3 is 3, because both are three units away from zero on the number line.

In equations, when you see something like \(|x|\), it means we're only interested in the magnitude of x, not its sign. This property is used in many algebraic manipulations. In the exercise, we used \(|2(-x)| = 2|x|\), since the absolute value of a negative number doesn't change its absolute value.
Function Transformation
In algebra, function transformations involve shifting, stretching, compressing, or reflecting the graph of a function. When working with functions, understanding how these transformations work makes it easier to manipulate and solve them.

To find \(g(-x)\) from \(g(x)\), we replaced every instance of x with -x in the function. This resulted in a reflection of the function's graph across the y-axis. However, in our specific example, \(g(x) = |2x| - x^2 + 3\), the function is symmetric, meaning \(g(x) = g(-x)\)

Such transformations are essential for understanding how changes to the input variable (x) affect the output.
Polynomial Simplification
Simplifying polynomials is a fundamental skill in algebra. It involves combining like terms and performing operations within the polynomial to make it more manageable.

In our exercise, after replacing x with -x in \(g(x) = |2x| - x^2 + 3\), we had to simplify the terms.
  • The absolute value term \(|2(-x)| = |2x| = 2|x|\).
  • The square term \((-x)^2 = x^2\).
These simplifications made it clear that \(g(-x) = 2|x| - x^2 + 3\) is the same as the original function. Recognizing and simplifying these types of polynomial terms is essential for solving algebraic equations efficiently.
Symmetry in Functions
Symmetry in functions helps to understand their behavior and graphing properties. A function is symmetric with respect to the y-axis if \(f(x) = f(-x)\) for all x in the function's domain.

In our example, \(g(x) = 2|x| - x^2 + 3\), after transforming and simplifying \(g(-x)\), we found that it equaled the original function. This indicates that \(g(x)\) is symmetric with respect to the y-axis.

This kind of symmetry simplifies the analysis of functions, as it tells us the graph of \(g(x)\) is a mirror image about the y-axis. Recognizing symmetry can also make it easier to solve equations and understand the nature of the function.

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Most popular questions from this chapter

Given \(f(x)=2 x+4\) and \(g(x)=x^{2}\) a. Find \((f \circ g)(x)\). b. Find \((g \circ f)(x)\). c. Is the operation of function composition commutative?

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a. Graph \(m(x)=\frac{1}{2} x-2\) for \(x \leq-2\). b. Graph \(n(x)=-x+1\) for \(x>-2\). c. Graph \(t(x)=\left\\{\begin{array}{ll}\frac{1}{2} x-2 & \text { for } x \leq-2 \\ -x+1 & \text { for } x>-2\end{array}\right.\)

A car starts from rest and accelerates to a speed of \(60 \mathrm{mph}\) in \(12 \mathrm{sec} .\) It travels \(60 \mathrm{mph}\) for \(1 \mathrm{~min}\) and then decelerates for 20 sec until it comes to rest. The speed of the car \(s(t)\) (in mph) at a time \(t\) (in sec) after the car begins motion can be modeled by: \(s(t)=\left\\{\begin{array}{cl}\frac{5}{12} t^{2} & \text { for } 0 \leq t \leq 12 \\ 60 & \text { for } 12
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