Problem 30 - College Algebra

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

Step 1: Check if the function is one-to-one

First, we need to ensure that the function \( f(x) = x^2 - 6x + 3 \) is one-to-one over the interval \([3, \infty)\). Since this is a quadratic function with a vertex at \( x = 3 \) (because the derivative \( f'(x) = 2x - 6 = 0 \) at \( x = 3 \)), and the parabola opens upwards, it is one-to-one on \([3, \infty)\).

Step 2: Set the function equal to \( y \)

Let \( y = x^2 - 6x + 3 \). Our goal is to express \( x \) in terms of \( y \).

Step 3: Complete the square

Rewrite the quadratic function in vertex form by completing the square:1. Start with the expression: \( x^2 - 6x + 3 \).2. Take half of the \(-6\) (coefficient of \( x \)), square it: \((-3)^2 = 9\).3. Add and subtract this square: \(x^2 - 6x + 9 - 9 + 3 = (x - 3)^2 - 6\).So, \( y = (x - 3)^2 - 6 \).

Step 4: Solve for \( x \) in terms of \( y \)

Rearrange the equation \( y = (x - 3)^2 - 6 \) to express \( x \):1. Add 6 to both sides: \( (x - 3)^2 = y + 6 \).2. Take the square root of both sides, and since \( x \geq 3 \), use the positive root: \( x - 3 = \sqrt{y + 6} \).3. Solve for \( x \): \( x = \sqrt{y + 6} + 3 \).

Step 5: Write the inverse function

Express the inverse function using the expression for \( x \):The inverse function is \( f^{-1}(y) = \sqrt{y + 6} + 3 \), for \( y \geq 3 \).

Key Concepts

One-to-One Function

In mathematics, a one-to-one function is a function where each output is determined by exactly one input. This is also referred to as an injective function. The main property of a one-to-one function is that different inputs always lead to different outputs.
Why is this important? For a function to have an inverse, it needs to be one-to-one on its domain. This means for every value of the function, we can reverse the process and get back to the input. In our exercise, we had to check that the function \( f(x) = x^2 - 6x + 3 \) is one-to-one. Given that this function is quadratic, which is not one-to-one over its entire set of inputs, we must restrict it to a specific interval where it maintains its one-to-one nature. Here, the interval \([3, \infty)\) ensures that the function is increasing and therefore one-to-one.

Quadratic Function

A quadratic function is a polynomial function of degree two, which can be generally written as \( f(x) = ax^2 + bx + c \). These functions are represented graphically by a parabola.
Parabolas have several key properties:

• They have an axis of symmetry, which divides the parabola into two mirror-image halves.
• The vertex, which is either the maximum or minimum point of the parabola, depending on whether it opens downwards or upwards.

In the given exercise, the quadratic function \( f(x) = x^2 - 6x + 3 \) forms a parabola opening upwards. The vertex of this parabola is located at \( x = 3 \). Because the parabola opens upwards, the function is strictly increasing on the interval \([3, \infty)\). This property allows us to find the inverse for this specific interval.

Vertex Form

Vertex form is a specific way of expressing a quadratic function, useful for understanding the properties of the parabola such as its vertex.
The vertex form of a quadratic function is written as \( f(x) = a(x-h)^2 + k \), where:

• \( (h, k) \) is the vertex of the parabola.
• \( a \) determines the width and the direction of the parabola. If \( a > 0 \), it opens upwards, and if \( a < 0 \), it opens downwards.

To convert a standard form quadratic function like \( x^2 - 6x + 3 \) into vertex form, completing the square is often used. In our example, the vertex form was found to be \( (x - 3)^2 - 6 \), which indicates that the vertex is at \( (3, -6) \). This conversion is helpful for solving the inverse function as it clarifies the transformations done to a simple \( x^2 \) term.

Completing the Square

Completing the square is a method used to convert a quadratic expression in standard form into vertex form. It is particularly useful for solving quadratic equations and finding the vertex of a parabola.
Here’s how it works:

• Take the quadratic expression, such as \( x^2 - 6x + 3 \).
• Identify the coefficient of the \( x \) term, divide it by two, and square the result. For \( -6 \), this gives \( (-3)^2 = 9 \).
• Add and subtract this square inside the expression: \( x^2 - 6x + 9 - 9 + 3 \).
• This rewrites the quadratic as \( (x-3)^2 - 6 \).*\

By completing the square, the quadratic function is easier to manipulate, particularly when rewriting it in vertex form or when finding its inverse. In the solution, we used this method to express \( y = (x - 3)^2 - 6 \), preparing it for inversion.