Problem 20 - 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 inverse value \( g^{-1}(5) \) is 1.

Step 1: Identify the function

The function given is \( g(x) = x^2 + 4x \) with \( x \geq -2 \). We are tasked with finding \( g^{-1}(5) \), which means we need to find \( x \) such that \( g(x) = 5 \).

Step 2: Set up the equation

To find the value of \( x \) that satisfies \( g(x) = 5 \), we set the equation \( x^2 + 4x = 5 \). This becomes our equation to solve for \( x \).

Step 3: Rearrange the equation

Rearrange \( x^2 + 4x = 5 \) to form a quadratic equation: \( x^2 + 4x - 5 = 0 \).

Step 4: Apply the quadratic formula

Use the quadratic formula to solve for \( x \): \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -5 \).

Step 5: Calculate the discriminant

Compute the discriminant: \( b^2 - 4ac = 4^2 - 4(1)(-5) = 16 + 20 = 36 \).

Step 6: Solve for x using the quadratic formula

Substitute the discriminant and other values into the quadratic formula: \( x = \frac{-4 \pm \sqrt{36}}{2} \). Simplify to \( x = \frac{-4 \pm 6}{2} \).

Step 7: Determine potential x values

The possible solutions for \( x \) are \( x = \frac{2}{2} = 1 \) and \( x = \frac{-10}{2} = -5 \).

Step 8: Check domain constraints

We are given that \( x \geq -2 \). Thus, \( x = -5 \) is not valid. The only valid solution is \( x = 1 \).

Key Concepts

Quadratic Equations

Understanding quadratic equations is an important step in solving many algebraic problems. A quadratic equation is a polynomial equation of the second degree, generally written in the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). The equation can take various forms, but its defining feature is the \(x^2\) term.

The graph of a quadratic equation is a parabola. It opens upwards if \(a > 0\) and downwards if \(a < 0\). Solving a quadratic equation means finding the values of \(x\) that satisfy the equation, also known as the roots of the equation.

• Quadratic equations can have zero, one, or two real roots.
• The nature of the roots depends on the discriminant \(b^2 - 4ac\).

Tip: Always check if the quadratic can be factored before moving to more complex solutions like the quadratic formula.

Domain Constraints

Domain constraints are conditions that are put on the variable \(x\) within a function. They determine the set of permissible inputs for which the function is defined. Inverting a function, like finding \(g^{-1}(5)\), often requires evaluating these constraints.

In the given exercise, the function is defined as \(g(x) = x^2 + 4x\) with the restriction \(x \geq -2\). This means the function only considers values of \(x\) that are greater than or equal to \(-2\). Therefore, any solution to our equation must respect this constraint.

• Domain constraints limit the valid solutions when finding inverse functions.
• Always verify solutions against domain conditions to ensure they are acceptable.

Checking the domain is crucial. In this exercise, one potential solution \(x = -5\) is invalid because it does not meet the constraint \(x \geq -2\). Only \(x = 1\), which satisfies this domain, is valid.

Quadratic Formula

The quadratic formula is a reliable method for solving any quadratic equation \(ax^2 + bx + c = 0\). The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the possible values of \(x\) by substituting the coefficients \(a\), \(b\), and \(c\) from the equation.

The term \(b^2 - 4ac\) under the square root is called the discriminant. It determines the nature of the roots:

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, there is one real root (or a repeated root).
• If negative, there are no real roots, only complex roots.

In our example, by using \(a = 1\), \(b = 4\), and \(c = -5\), we found that the discriminant is 36, which is positive. This indicates two possible real roots. After calculating, we checked which root falls within the domain constraint of \(x \geq -2\) to find the valid solution \(x = 1\).