Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, find the inverse of the functions. $$ f(x)=x^{2}-6 x+3,[3, \infty) $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = 3 + \sqrt{x + 6} \).

Step 1: Recognize the Function Type

The given function is a quadratic function of the form \( f(x) = ax^2 + bx + c \) where \( a = 1 \), \( b = -6 \), and \( c = 3 \). Quadratic functions can only have inverses over restricted domains where the function is either strictly increasing or decreasing. The domain \( [3, \infty) \) ensures the function is strictly increasing.

Step 2: Write the Equation for Inversion

Set \( y = f(x) = x^2 - 6x + 3 \). We need to solve this equation for \( x \) in terms of \( y \). This means we want to express \( x = g(y) \), where \( g \) is the inverse function.

Step 3: Solve the Quadratic Equation

Rewrite \( y = x^2 - 6x + 3 \) as \( x^2 - 6x + (3 - y) = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \). Here, \( a = 1, b = -6, \) and \( c = 3 - y \).

Step 4: Apply the Quadratic Formula

Substitute into the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (3-y)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 12 + 4y}}{2} = \frac{6 \pm \sqrt{4y + 24}}{2}. \] Since the function is increasing on \([3, \infty)\), take the positive solution: \[ x = \frac{6 + \sqrt{4y + 24}}{2}. \]

Step 5: Simplify the Expression

Simplify the expression found: \[ x = \frac{6 + \sqrt{4(y + 6)}}{2} = 3 + \sqrt{y + 6}. \] This gives the inverse function \( g(y) = 3 + \sqrt{y + 6} \).

Step 6: Write the Inverse Function

The inverse of the function \( f(x) = x^2 - 6x + 3 \) is \( f^{-1}(x) = 3 + \sqrt{x + 6} \). The domain of this inverse function is determined by the range of the original function \( f \), which is \([3, \infty)\).

Key Concepts

Quadratic Functions

Quadratic functions are one of the simplest types of polynomial functions, characterized by the highest degree being two. They are often expressed in the standard form: \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The graph of a quadratic function is called a parabola, which can open upwards if \( a > 0 \) or downwards if \( a < 0 \).

One of the main features of quadratic functions is their vertex, which is the highest or lowest point on the graph, depending on the parabola's orientation. The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola.

Quadratic functions are crucial in various real-world applications, including physics and engineering, as they can model trajectories, area problems, and more.

Domain and Range

Understanding the domain and range of a function is fundamental in mathematics. The domain of a function is the complete set of possible values of the independent variable, usually \( x \). For quadratic functions, the domain is typically all real numbers if there are no restrictions implied by the context of a problem.

The range, on the other hand, refers to the set of possible output values (or \( y \)-values) of a function. For a quadratic function expressed as \( f(x) = ax^2 + bx + c \), the range is determined by the direction in which the parabola opens and the vertex.

In the exercise, the quadratic function \( f(x) = x^2 - 6x + 3 \) is limited to the domain \([3, \infty)\), forcing the function to be strictly increasing. This restriction simplifies finding the inverse, ensuring that each output corresponds to exactly one input, which is necessary for a function to have an inverse.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations in which the highest exponent of the variable is two. The general quadratic equation is written as \( ax^2 + bx + c = 0 \), and the solutions for \( x \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides the values of \( x \) where the parabola intersects the \( x \)-axis (i.e., the roots of the equation). The "\( \pm \)" sign indicates that there can be two solutions.

• The discriminant \( b^2 - 4ac \) found under the square root determines the nature of the roots. If positive, there are two real roots; if zero, there is one real root; and if negative, the roots are complex numbers.

In the function inversion process in our exercise, we substitute known values into the quadratic formula to solve for \( x \, \), securing an expression for the inverse function.

Function Inversion

Function inversion is a process that "reverses" the original function, swapping its inputs and outputs. To find an inverse, we solve the equation \( y = f(x) \) for \( x \) in terms of \( y \) and denote the inverse function as \( f^{-1}(x) \).

Inverting a function means finding out how to "undo" the effect of a given function. For the function \( f(x) = x^2 - 6x + 3 \), finding the inverse required using the quadratic formula and taking advantage of the restricted domain.

• This ensures that the function is one-to-one, which is a requirement for functions to have an inverse.
• After finding \( x \) in terms of \( y \) and simplifying, the inverse function was determined to be \( f^{-1}(x) = 3 + \sqrt{x + 6} \).

In practical terms, this inverse function allows us to trace back from the output to the input, maintaining mathematical relationships in data processing, coding, and other analytical areas.