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Chapter 2: Problem 22

Find the inverse function \(f^{-1}\) and state its domain. $$f(x)=x^{2}+2 x, x \geqslant-1$$

Short Answer

Expert verified
The inverse function is \( f^{-1}(x) = -1 + \sqrt{1+x} \) and its domain is \( x \geq -1 \).

Step by step solution

01

Set Up the Equation for Inversion

Start by replacing the function notation with a variable to make the equation easier to work with. Let \( y = f(x) = x^2 + 2x \). The task is to find \( x \) in terms of \( y \), since \( x \) will be the function of the inverse.
02

Solve for x

The given function is \( y = x^2 + 2x \). Rearranging terms gives \( x^2 + 2x - y = 0 \). This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a=1 \), \( b=2 \), and \( c=-y \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \).
03

Apply the Quadratic Formula

Using the quadratic formula for \( x^2 + 2x - y = 0 \):\[x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} \]Simplifying gives:\[x = \frac{-2 \pm \sqrt{4 + 4y}}{2} = \frac{-2 \pm \sqrt{4(1+y)}}{2} = \frac{-2 \pm 2\sqrt{1+y}}{2} = -1 \pm \sqrt{1+y}\]
04

Choose the Correct Solution

Since \( x \geq -1 \), we select \( x = -1 + \sqrt{1+y} \) as the inverse because \( x = -1 - \sqrt{1+y} \) would not satisfy the given domain of the original function.
05

State the Inverse Function

The inverse function is given by \( f^{-1}(y) = -1 + \sqrt{1+y} \). By replacing \( y \) with \( x \), the usual representation is \( f^{-1}(x) = -1 + \sqrt{1+x} \).
06

Determine the Domain of the Inverse Function

Recall that the original function has the range \( y \geq -1 \). Thus, the inverse function's domain must reflect this range. Therefore, the domain of \( f^{-1} \, \) is \( x \geq -1 \).

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Key Concepts

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

Quadratic Equation
A quadratic equation is a type of polynomial equation that takes the standard form: \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero.Key Features:
  • The highest exponent of the variable \( x \) is 2, which defines its name "quadratic."
  • The solutions of a quadratic equation are often found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • The expression under the square root sign, \( b^2 - 4ac \), is known as the discriminant. It determines the nature of the roots of the quadratic equation:
    • If the discriminant is positive, the equation has two distinct real roots.
    • If it is zero, there is one real root (a repeated root).
    • If it is negative, the equation has no real roots, but two complex roots.
In the context of finding an inverse function, we treat the function as a quadratic equation in terms of \( x \). By rewriting and manipulating the original function, students can solve for \( x \) using the quadratic formula, achieving an understanding of inverse functions in relation to quadratic functions.
Domain of a Function
The domain of a function refers to all possible input values (often \( x \)) which allow the function to work. For a function to have an inverse, it must be one-to-one, which in some cases requires restricting its domain.Understanding the Domain:
  • To determine the domain of a function, consider any limitations, such as division by zero or the square root of a negative number, which would make the expression undefined.
  • The original function given, \( f(x) = x^2 + 2x \), specifies that \( x \geq -1 \). This restriction ensures the function is one-to-one so that an inverse can exist.
When looking at the inverse function, its domain is determined by the range of the original function. The inverse function found is \( f^{-1}(x) = -1 + \sqrt{1+x} \), having a domain of \( x \geq -1 \), since the original function \( f(x) \) has the range \( y \geq -1 \). By confirming these domains, students can better understand how to set parameter boundaries for functions and their inverses.
Function Range
The range of a function includes all possible output values (often \( y \)) that result when the function is evaluated across its domain. Understanding the range is key to defining the domain of the inverse function.Determining the Range:
  • The range is essentially "the spread" of the possible outputs of the function when you plug in every element from the domain.
  • For the quadratic function \( f(x) = x^2 + 2x \) with the domain \( x \geq -1 \), the lowest point, or the vertex, provides the minimum value of \( y \), which stems from completing the square or analyzing the parabola's vertex.
    • Here, the vertex occurs at \( x = -1 \), and substituting \( x = -1 \) yields \( y = -1 \). Thus, the range starts at \( y = -1 \) and extends to infinity.
Understanding the range is crucial because it directly transitions to the domain of the inverse function. When finding the inverse of \( f(x) \), the range \( y \geq -1 \) is used to set the domain \( x \geq -1 \) for the inverse function \( f^{-1}(x) \). Recognizing this relationship helps ensure students are adequately grasping how inverses map between domains and ranges.

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Most popular questions from this chapter

For each pair of functions,\( find \)(f \circ g)(x)\( and \)(g \circ f)(x)$ and state the domain for each. $$f(x)=3 x+5, g(x)=\frac{x-5}{3}$$

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