Problem 16 - Algebra 2 and Trigonometry

Question

In \(11-16,\) determine if the function has an inverse. If so, list the pairs of the inverse function. If not, explain why there is no inverse function. $$ \left\\{(x, y) : y=x^{2}+2 \text { for } 0 \leq x \leq 5\right\\} $$

Step-by-Step Solution

Answer

The function has an inverse, given by \( f^{-1}(y) = \sqrt{y - 2} \), valid for \( y \geq 2 \).

Step 1: Understand the Function

The given function is described by the set of points \( (x, y) \) where \( y = x^2 + 2 \), and \( 0 \leq x \leq 5 \). This describes a function in the form of a quadratic equation, which generally forms a parabola.

Step 2: Check for One-to-One Property

For a function to have an inverse, it must be one-to-one (bijective). A quadratic function is one-to-one over a restricted domain where the function is either strictly increasing or strictly decreasing. Since the function \( y = x^2 + 2 \) with \( 0 \leq x \leq 5 \) is increasing (as the derivative \( \frac{d}{dx}(x^2) = 2x \geq 0 \) for \( x \geq 0 \)), it is one-to-one in this interval.

Step 3: Determine the Inverse Function

To find the inverse, solve for \( x \) in terms of \( y \). Set \( y = x^2 + 2 \). Re-arrange the equation to express \( x \) in terms of \( y \):\[ y - 2 = x^2 \] Take the square root of both sides:\[ x = \sqrt{y - 2} \] Since \( x \geq 0 \) (as given by the domain \( 0 \leq x \leq 5 \)), we only consider the positive square root.

Step 4: Identify Inverse Pairs

Given the inverse formula \( x = \sqrt{y - 2} \), let's express \( f^{-1} \) as \( f^{-1}(y) = \sqrt{y - 2} \). - When \( y = x^2 + 2 \), it aligns with \( x = \sqrt{y - 2} \) such that for every pair \((x, y)\) in \( f \), we have \((y, x)\) in \( f^{-1} \). - A few example pairs are: - \((0, 2)\) gives \( (2, 0) \) - \((4, 18)\) gives \((18, 4)\)

Key Concepts

Quadratic Functions

Quadratic functions are equations that can be expressed in the form \( y = ax^2 + bx + c \). They are characterized by their U-shaped graphs, known as parabolas. These functions may open upward or downward based on the sign of the leading coefficient \( a \). If \( a > 0\), the parabola opens upward, whereas \( a < 0 \) opens it downward. In this exercise, the function \( y = x^2 + 2 \) is a simple quadratic with \( a = 1 \), so it opens upwards.

There are several key features of quadratic functions worth noting:

• The vertex is the highest or lowest point on the graph. In our function, the vertex is at the point where \( x = 0 \), yielding \( y = 2 \).
• The axis of symmetry, a vertical line through the vertex, divides the parabola into two mirror-image halves. For the function given, this line is \( x = 0 \).

One-to-One Property

The one-to-one property of a function ensures that each input has a unique output, and vice versa. For a function to have an inverse, it must be one-to-one. Quadratic functions are not inherently one-to-one because they typically cover values in both positive and negative directions.

However, by restricting the domain, we can make a quadratic function one-to-one. In this case, the domain is limited to \( 0 \leq x \leq 5 \), making the function \( y = x^2 + 2 \) one-to-one over this interval. Here, the function is strictly increasing because we only consider \( x \geq 0 \). This allows each \( y \) value to have a unique corresponding \( x \) value, enabling the possibility of an inverse.

Bijective Functions

A bijective function is one that is both injective (one-to-one) and surjective (onto). When a function satisfies both these conditions, it means every \( y \) value in the range has exactly one corresponding \( x \) value in the domain.

In this exercise, the function \( y = x^2 + 2 \) defined for \( 0 \leq x \leq 5 \) is injective because of the restricted domain. Each \( x \) maps to a unique \( y \), satisfying the injective condition. The function is also surjective onto its range \( [2, 27] \), meaning that for every \( y \) in this range, there exists an \( x \) such that \( y = x^2 + 2 \). Hence, this function is bijective over the given domain and range.

Function Domain and Range

The domain of a function is the set of all possible input values \( x \), while the range is the set of all possible output values \( y \). Identifying these is crucial for understanding how a function behaves.

For the function \( y = x^2 + 2 \) with \( 0 \leq x \leq 5 \), the domain is straightforwardly \([0, 5]\), which means that the input \( x \) can take any value from 0 to 5, inclusive. The range is determined by the behavior of the quadratic expression. With \( y = x^2 + 2 \), when \( x = 0 \), \( y = 2 \), and when \( x = 5 \), \( y = 27 \). Therefore, the range is \([2, 27]\), encompassing all possible output values between these limits. This understanding is crucial when determining if a function has an inverse, as it helps to ensure the inverse function properly maps from range back to domain.