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Chapter 2: Problem 118

If \(f\) is an even function defined on the interval \([-5,5]\), then the real values of \(x\) satisfying the equation $$ f(x)=f\left(\frac{x+1}{x+2}\right) \text { are } $$ (A) \(\frac{-1 \pm \sqrt{5}}{2}\) (B) \(\frac{-3 \pm \sqrt{5}}{2}\) (C) \(\frac{-2 \pm \sqrt{5}}{2}\) (D) None of these

Short Answer

Expert verified
(B) \(\frac{-3 \pm \sqrt{5}}{2}\).

Step by step solution

01

Understanding Even Functions

An even function is symmetric about the y-axis, meaning that for every value x in its domain, \(f(x) = f(-x)\). Given that f is even and defined on \([-5,5]\), this property will be crucial in solving the problem.
02

Setting Up the Equation

We need to solve \(f(x) = f\left(\frac{x+1}{x+2}\right)\). Since \(f\) is an even function, \(f(a) = f(b)\) implies either \(a = b\) or \(a = -b\). Apply this property to \(x\) and \(\frac{x+1}{x+2}\).
03

Solving for Equality

Start by solving for \(x = \frac{x+1}{x+2}\). Multiply both sides by \(x+2\) and simplify to find: \(x(x+2) = x+1\). Rearrange to obtain the quadratic equation \(x^2 + x - 2 = 0\).
04

Factoring the Quadratic

The quadratic equation \(x^2 + x - 2 = 0\) can be factored as \((x-1)(x+2) = 0\). This gives the solutions \(x = 1\) and \(x = -2\).
05

Solving for Opposites

Next, consider \(x = -\frac{x+1}{x+2}\). Again, multiply by \(x+2\) to get \(x(x+2) = -(x+1)\). This simplifies to \(x^2 + 3x + 1 = 0\).
06

Solving the Second Quadratic

Using the quadratic formula for \(x^2 + 3x + 1 = 0\), we find \(x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}\). Simplify to get roots \(x = \frac{-3 \pm \sqrt{5}}{2}\).
07

Conclusion and Answer Selection

The real values of \(x\) satisfying the original equation include the solutions from \(x^2 + 3x + 1 = 0\), which are \(x = \frac{-3 \pm \sqrt{5}}{2}\). Therefore, the correct answer is (B).

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Key Concepts

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

Symmetry about y-axis
An even function has a distinct characteristic: it is symmetric about the y-axis. This means the function mirrors itself across the y-axis. For an even function, the equation \( f(x) = f(-x) \) holds true for every \( x \) within its domain.
When visualizing a graph, imagine plotting a point for \( f(x) \) and then see a corresponding point at \( f(-x) \) directly opposite the y-axis. This reflection property is crucial for solving equations involving even functions.
  • For example, if \( f(2) = 5 \), then \( f(-2) \) must also equal 5.
  • This property helps in identifying solutions because the value of \( f \) is purely based on the magnitude of \( x \), not its sign.
This symmetry plays a vital role in complex algebraic manipulations, helping identify relationships and solutions in equations like the one given in this exercise.
Quadratic Equations
Quadratic equations are a common type of polynomial equation represented in the form \( ax^2 + bx + c = 0 \). In this exercise, we encountered two quadratic equations: \( x^2 + x - 2 = 0 \) and \( x^2 + 3x + 1 = 0 \).
To solve quadratic equations, we often use:
  • Factoring: If the quadratic can easily be rewritten as a product of two binomials.
  • Completing the square: This method involves rearranging the equation into a perfect square.
  • Quadratic formula: For any quadratic \( ax^2 + bx + c = 0 \), solutions are given by the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
The discriminant \( b^2 - 4ac \) within the quadratic formula determines the nature of the roots:
  • If positive, there are two distinct real roots.
  • If zero, the roots are real and equal.
  • If negative, the roots are complex.
Understanding these methods aids in solving broader mathematical problems and finding specific values in polynomial equations.
Domain of Functions
The domain of a function specifies all possible input values \( x \) that the function can accept. For an even function defined on \([-5,5]\), the domain is restricted within this interval.
Knowing the domain is critical because it imposes limitations on the solutions you seek. If a solution falls outside the domain, it becomes invalid for that particular function.
  • Considerations about domain help prevent extraneous solutions from factoring into final answers.
  • This is especially important when dealing with rational functions or expressions whereby certain values might make the function undefined (like division by zero).
In our original exercise, we focus only on values within the given domain, ensuring any solution of the equation \( f(x) = f\left(\frac{x+1}{x+2}\right) \) must comply with this range. Thus, domain considerations act as a filter for practical and accurate solutions in real-world scenarios.

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Most popular questions from this chapter

If the function \(f:[1, \infty) \rightarrow[1, \infty)\) is defined by \(f(x)=\) \(2^{x(x-1)}\), then (A) \(f\) is one-one (B) \(f\) is onto (C) \(f^{-1}(x)=\frac{1+\sqrt{1+4 \log _{2} x}}{2}\) (D) \(f^{-1}(x)=\frac{1-\sqrt{1+4 \log _{2} x}}{2}\)

If \(f: R \rightarrow S\), defined by \(f(x)=\sin x-\sqrt{3} \cos x+1\), is onto, then the interval of \(S\) is (A) \([0,3]\) (B) \([-1,1]\) (C) \([0,1]\) (D) \([-1,3]\)

If \(f:(0, \pi) \rightarrow R\) be defined by \(f(x)=\sum_{k=1}^{n}([1+\sin k x])\), where \([x]\) denotes the integral part of \(x\), then the range of \(f(x)\) is (A) \(\\{n-1, n+1\\}\) (B) \(\\{n-1, n, n+1\\}\) (C) \(\\{n, n+1\\}\) (D) None of these

Let \(f: R \rightarrow R\) be a function defined by, \(f(x)=\) \(-\frac{|x|^{3}+|x|}{1+x^{2}}\), then the graph of \(f(x)\) lies in which quadrant \((s) ?\) (A) I and II (B) I and III (C) II and III (D) III and IV

The function $$ f(x)=\sin ^{-1}\left(x-x^{2}\right)+\sqrt{1-\frac{1}{|x|}}+\frac{1}{\left[x^{2}-1\right]} $$ is defined in the interval (where \([\cdot]\) is the greatest integer) (A) \(x \in\left(\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\) (B) \(x \in\left(1, \frac{1+\sqrt{5}}{2}\right)\) (C) \(x \in\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]\) (D) \(x \in\left(-\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\)
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