Problem 24 - Algebra and Trigonometry

Question

If \(g(x)=x^{2}+4 x\) with \(x \geq-2,\) find \(g^{-1}(5)\)

Step-by-Step Solution

Answer

The value of \(g^{-1}(5)\) is 1.

Step 1: Understand the Inverse Function

To find the inverse function, we need to interchange the roles of \(x\) and \(y\), and then solve for \(x\). So, if \(y = g(x) = x^2 + 4x\), we need to express \(x\) in terms of \(y\).

Step 2: Solve for \(x\) in terms of \(y\)

Rearrange the equation \( y = x^2 + 4x \) to solve for \(x\). First rewrite it as \( x^2 + 4x - y = 0 \). This is a quadratic equation in \(x\).

Step 3: Apply the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(x\). Here, \(a = 1\), \(b = 4\), and \(c = -y\). Substitute these values to get:\[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-y)}}{2(1)} \]\[ x = \frac{-4 \pm \sqrt{16 + 4y}}{2} \]\[ x = \frac{-4 \pm \sqrt{4(y + 4)}}{2} \]\[ x = \frac{-4 \pm 2\sqrt{y + 4}}{2} \]\[ x = -2 \pm \sqrt{y + 4} \]

Step 4: Consider the Domain

Since \(x \geq -2\), we select \(x = -2 + \sqrt{y + 4}\) as the inverse function, \(g^{-1}(y)\). This ensures that the selected branch satisfies the domain constraint.

Step 5: Find \(g^{-1}(5)\)

Substitute \(y = 5\) into the inverse function we found:\[ x = -2 + \sqrt{5 + 4} \]\[ x = -2 + \sqrt{9} \]\[ x = -2 + 3 \]\[ x = 1 \]

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool in solving quadratic equations. Quadratic equations often appear in the form of \( ax^2 + bx + c = 0 \). To find the roots of such equations, we rely on the quadratic formula, which is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation, where \( a eq 0 \). The symbol \( \pm \) indicates that there are generally two solutions: one involving addition and the other subtraction. Using the formula, we can determine the exact points at which the parabola, represented by the quadratic equation, intersects the x-axis. This is useful for finding solutions for inverse functions, as it helps to rearrange equations and solve for unknowns. For instance, when dealing with the function \( g(x) = x^2 + 4x \), the quadratic formula aids in expressing \( x \) in terms of \( y \).

Domain of a Function

The domain of a function refers to the set of all possible input values \( x \) that the function can accept without leading to undefined or non-real outcomes. It’s essential to consider the domain when discussing functions because it defines where the function is applicable and determines the context in which solutions are valid. For the function \( g(x) = x^2 + 4x \), given the domain indicator \( x \geq -2 \), we need to remain within this constraint to ensure that our solutions are consistent and meaningful. This is particularly relevant when defining inverse functions, as the domain of the original function impacts the range (possible output values) and thus influences which part of the inverse function is correct. In this case, while solving for the inverse, we choose \( x = -2 + \sqrt{y + 4} \), ensuring that \( x \geq -2 \), maintaining the function's domain integrity.

Solving Quadratic Equations

Solving quadratic equations is an important skill when working with polynomials, as it allows us to find the values of \( x \) that satisfy the equation. There are several methods to solve quadratic equations:

• Factoring: When a quadratic can be expressed as a product of two binomials.
• Completing the Square: Rewriting the equation as a perfect square trinomial.
• Quadratic Formula: A more universally applicable method, ideal when other methods are cumbersome.

In our specific problem, \( x^2 + 4x - y = 0 \), we used the quadratic formula because it provides a structured way to handle the inverse function problem. By identifying \( a = 1 \), \( b = 4 \), and \( c = -y \), we were able to substitute these directly into the quadratic formula. This process allowed us to derive the inverse function \( g^{-1}(y) \) by solving for \( x \) in terms of \( y \). Being adept at solving quadratics ensures you can tackle a wide range of mathematical challenges efficiently.