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Let \(f(x)=a x^{2}+b x+c, a>0 .\) Find \(f^{-1}\) if the domain of \(f\) is restricted to $$ \text { (a) } x \geq-b /(2 a) \quad \text { (b) } x \leq-b /(2 a) $$

Short Answer

Expert verified
The inverse is two parts: for \(x \geq -\frac{b}{2a}\), \(f^{-1}(y) = -\frac{b}{2a} + \sqrt{\frac{y-c}{a}}\), and for \(x \leq -\frac{b}{2a}\), \(f^{-1}(y) = -\frac{b}{2a} - \sqrt{\frac{y-c}{a}}\).

Step by step solution

01

Identify the function and its form

The given function is a quadratic function: \(f(x) = ax^2 + bx + c\), and it is given that \(a > 0\). This means \(f(x)\) is a parabola that opens upwards. We are required to find the inverse of this function when its domain is restricted.
02

Understanding the vertex of the parabola

The vertex of the parabola \(f(x) = ax^2 + bx + c\) is found at \(x = -\frac{b}{2a}\). Knowing this will help in understanding the behavior of the function for given domain restrictions.
03

Finding inverse for \(x \geq -\frac{b}{2a}\)

For \(x \geq -\frac{b}{2a}\), the function is one-to-one as \(f(x)\) is increasing on this interval. Start by expressing \(y = ax^2 + bx + c\), then rearrange to solve for \(x\):- Rearrange and isolate the terms: \[ y - c = ax^2 + bx \]- Complete the square: \[ y - c = a(x^2 + \frac{b}{a}x) \] Add and subtract \((\frac{b}{2a})^2\) inside the bracket: \[ y - c = a((x + \frac{b}{2a})^2 - (\frac{b}{2a})^2) \] Simplify to: \[ (x + \frac{b}{2a})^2 = \frac{y - c}{a} + (\frac{b}{2a})^2 \] Solve for \(x\): \[ x = -\frac{b}{2a} + \sqrt{\frac{y-c}{a}} \]
04

Finding inverse for \(x \leq -\frac{b}{2a}\)

For \(x \leq -\frac{b}{2a}\), \(f(x)\) is decreasing since \(x\) is on the left side of the vertex where the parabola opens upwards. Start by expressing \(y = ax^2 + bx + c\) and follow similar steps to previous part:- Solve the same quadratic equation: \[ (x + \frac{b}{2a})^2 = \frac{y - c}{a} + (\frac{b}{2a})^2 \] - For the decreasing portion, it means taking the negative square root: \[ x = -\frac{b}{2a} - \sqrt{\frac{y-c}{a}} \]
05

Write the inverse functions

With the previous steps done, we have:For \(x \geq -\frac{b}{2a}\), the inverse is:\[ f^{-1}(y) = -\frac{b}{2a} + \sqrt{\frac{y-c}{a}} \]For \(x \leq -\frac{b}{2a}\), the inverse is:\[ f^{-1}(y) = -\frac{b}{2a} - \sqrt{\frac{y-c}{a}} \]

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Key Concepts

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

Quadratic Function
Quadratic functions are a type of polynomial function that are represented by the equation \(f(x) = ax^2 + bx + c\). This is called the standard form of a quadratic function. Here, \(a\), \(b\), and \(c\) are constants, with \(aeq0\). The graph of a quadratic function is a curve known as a parabola. This parabola can open upwards or downwards depending on the sign of \(a\).
Since \(a > 0\) for this exercise, the parabola opens upwards. The basic structure of a quadratic function makes it easy to solve or "invert" for different domains, as long as it remains one-to-one. This exercise involves finding the inverse function by restricting the domain of the original function to ensure it meets the criteria for inversion. Quadratic functions play a fundamental role in various scientific, engineering, and business applications and are essential in understanding more complex mathematical models.
When dealing with quadratic functions and their inverses, it is crucial to understand how their shapes and orientations change based on their coefficients and domain restrictions.
Parabola Vertex
The parabola vertex is a special point that gives valuable information about the quadratic function. It is the maximum or minimum point of the parabola depending on whether it opens downward or upward, respectively. For a quadratic function in standard form \(f(x) = ax^2 + bx + c\), the vertex is located at \(x = -\frac{b}{2a}\). Knowing the vertex helps in understanding the behavior of the parabola and its symmetry.
In the context of this exercise, the vertex divides the parabola into two distinct parts. This division simplifies the process of finding the inverse by treating each segment separately. When restricted to \(x \geq -\frac{b}{2a}\), the parabola only includes values where it increases. Conversely, if the domain is \(x \leq -\frac{b}{2a}\), the parabola is decreasing. Comprehending the vertex aids in determining the function's increasing or decreasing nature over specific domains.
Utilizing the vertex formula allows for an easier way to express and manipulate the quadratic equation during methods like completing the square, ultimately assisting in finding accurate inverse functions.
Completing the Square
Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial, which simplifies the equation for solving, graphing, or inverting. It is especially significant when finding the inverse of a quadratic function. This transformation involves changing \(ax^2 + bx + c \) into something like \(a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola.
In the exercise, completing the square is pivotal in rearranging the quadratic equation for both restricted domains \(x \geq -\frac{b}{2a}\) and \(x \leq -\frac{b}{2a}\). By rewriting the original function's quadratic term into a perfect square, solving for \(x\) becomes more straightforward. Here's a simplified breakdown of the steps:
  • Start with \(y = ax^2 + bx + c\).
  • Rearrange into \(y-c = a(x^2 + \frac{b}{a}x)\).
  • Add and subtract \((\frac{b}{2a})^2\) inside the parentheses to complete the square.
  • The equation now looks like \((x + \frac{b}{2a})^2 = \frac{y - c}{a} + (\frac{b}{2a})^2\).
Applying this method enables us to isolate \(x\) more easily and find the inverse function, making it a valuable tool for many algebraic applications, including working with quadratic functions.

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Most popular questions from this chapter

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