Problem 28 - Algebra 2 and Trigonometry

Question

When the function \(g(x)=x^{2}+8 x+18\) is restricted to the interval \(x \geq 2,\) the inverse is \(g^{-1}(x)=\sqrt{x-2}-4\) a. Graph g for values of \(x \geq 2 .\) Graph \(\mathrm{g}^{-1}\) on the same set of axes. b. What is the domain of \(\mathrm{g}\) ? What is the range of \(\mathrm{g}\) ? c. What is the domain of \(\mathrm{g}^{-1}\) ? What is the range of \(\mathrm{g}^{-1}\) ? d. Describe the relationship between the domain and range of \(\mathrm{g}\) and its inverse.

Step-by-Step Solution

Answer

The domain of \(g\) is \([2, \infty)\), and the range is \([30, \infty)\). The domain of \(g^{-1}\) is \([30, \infty)\), and the range is \([2, \infty)\). These swap for the inverse.

Step 1: Graph g(x) and its inverse

To graph both functions, begin with identifying the function. For \(g(x) = x^2 + 8x + 18\), restrict the domain to \(x \geq 2\). This parabola opens upwards. \(g^{-1}(x) = \sqrt{x-2}-4\) is a square root function horizontally shifted and reflected. The range for \(g\) starts from the value of \(g(2)\) upwards, so compute it, yielding \(g(2) = 2^2 + 8 \cdot 2 + 18 = 30\). Thus, the range starts at 30. Draw the parabola \(g\) starting at \(x = 2\) with \(g(2)=30\) as its vertex upwards and overlay its inverse as a reflection over the line \(y = x\).

Step 2: Identify the domain and range of g(x)

Since \(g(x) = x^2 + 8x + 18\) is defined for \(x \geq 2\), the domain of \(g\) is \([2, \infty)\). As \(x\) grows, \(g(x)\) increases because it is a parabolic function with a minimum vertex at \(x=2\). Therefore, the range of \(g\) is \([30, \infty)\).

Step 3: Identify the domain and range of g^{-1}(x)

The domain of \(g^{-1}(x) = \sqrt{x-2} - 4\) is found by ensuring the expression under the square root is non-negative, so \(x-2 \geq 0\), or \(x \geq 2\). But it is derived from the range of \(g\), so the domain is actually \([30, \infty)\). For the range, since it represents the \(x\) values of \(g\), it starts from \(g(2) = 30\), so it is \([2, \infty)\), as \(g^{-1}\) returns input values into the original domain.

Step 4: Analyze the relationship between g and g^{-1}

The domain of \(g\) is \([2, \infty)\) and its range is \([30, \infty)\), representing the values \(g\) can output. The inverse \(g^{-1}\) reverses these roles: its domain is \([30, \infty)\) while its range is \([2, \infty)\). Hence, the domain of \(g\) becomes the range of \(g^{-1}\) and vice versa.

Key Concepts

Domain and Range

Understanding the domain and range of a function is crucial in mathematics. The domain of a function is the complete set of possible input values (usually represented as \(x\) values). For the function \(g(x) = x^2 + 8x + 18\) with the restriction \(x \geq 2\), the domain is all \(x\) values greater than or equal to 2. This is because the given condition constrains \(x\) to not be less than 2.

Meanwhile, the range of a function is the complete set of possible output values (or \(y\) values). The range of \(g(x)\) is found by calculating \(g(2)\) which equals 30. As the parabola moves upwards from \(x=2\), the range becomes \([30, \infty)\). Therefore, the smallest \(y\) value is 30 and it increases infinitely as \(x\) increases.

Inverse functions switch the roles of domain and range. For \(g^{-1}(x)\), the domain, derived from \(g's\) range, starts at 30 and the range corresponds to \(g's\) original domain, \([2, \infty)\). This illustrates how inputs and outputs are interchanged in inverse functions.

Quadratic Functions

Quadratic functions are fundamental in algebra and are characterized by their general form \( ax^2 + bx + c \). The function \( g(x) = x^2 + 8x + 18 \) is a classic example with \(a = 1\), \(b = 8\), and \(c = 18\). These functions represent parabolas, and depending on the coefficient \(a\), the parabola can open upwards (\(a > 0\)) or downwards (\(a < 0\)).

In our example, since \(a = 1\), the parabolic graph of \(g(x)\) opens upwards, indicating that the vertex represents the minimum point of the curve. The vertex can be calculated using the formula \(x = -\frac{b}{2a}\). Here, the calculation gives \(x = -4\), but since we restrict \(x\) to \(x \geq 2\), we focus on this section of the curve. At \(x = 2\), \(g(x)\) is at its lowest on the graph with a value of 30, hence, acting as the beginning of the restricted function's range. This idea of constraining the domain highlights an important aspect of utilizing and graphing quadratic functions, especially when determining their behaviors over specific intervals.

Parabola Graphing

Graphing parabolas involves interpreting the quadratic function and plotting its curve on a graph. For the function \(g(x) = x^2 + 8x + 18\) within \(x \geq 2\), begin by determining key features like the vertex, axis of symmetry, and direction of opening. With \(g(x)\), since \(a = 1\), the parabola opens upwards.

The vertex, critical for graphing, occurs at \(x = -4\), but due to the restriction, graphing commences from \(x=2\) upwards. At this point, the computed \(g(2)\) value of 30 is the \(y\)-intercept vertex of the restricted graph section.

When plotting inverse functions like \(g^{-1}(x) = \sqrt{x-2} - 4\), start by understanding that it mirrors the graph of \(g(x)\) across the line \(y = x\). This inverse graph begins at \(x=30\) and rises. Ensure that both graphs reflect each other accurately over the \(y = x\) line by checking corresponding points. Graphing inverses requires flipping the domain and range of the original function, underlining the relationship between inverse functions and their originals.