Problem 53 - Algebra and Trigonometry

Question

Find the inverse function of \(f\) $$ f(x)=4-x^{2}, \quad x \geq 0 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \sqrt{4 - x} \), with domain \( 0 \leq x \leq 4 \).

Step 1: Define the function and replace

First, we identify the original function: \( f(x) = 4 - x^2 \) with the condition \( x \geq 0 \). To find the inverse, we start by replacing \( f(x) \) with \( y \). Thus, we have the equation \( y = 4 - x^2 \).

Step 2: Swap variables

To find the inverse function, swap \( x \) and \( y \) in the equation from Step 1. This gives us \( x = 4 - y^2 \). The next step is to solve this equation for \( y \).

Step 3: Solve for y

Rearrange the equation from Step 2 to isolate \( y^2 \). This gives us \( y^2 = 4 - x \). Next, take the square root of both sides to solve for \( y \). Since \( x \geq 0 \), we take the non-negative root: \( y = \sqrt{4 - x} \).

Step 4: Define the inverse function

The function \( f^{-1}(x) \) is defined by \( y \). Therefore, the inverse function of \( f \) is \( f^{-1}(x) = \sqrt{4 - x} \). Note that the domain of this inverse function is \( 0 \leq x \leq 4 \), due to the condition \( x \geq 0 \) in the original function.

Key Concepts

Quadratic Functions

Quadratic functions are polynomial functions of degree 2. The general form of a quadratic function is given by \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These functions graph into a shape called a parabola.

• The term \( ax^2 \) is the quadratic term, and it's the most crucial part because it determines the shape and direction of the parabola. If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
• In the exercise, the function \( f(x) = 4 - x^2 \) is a form of a downward opening parabola, indicating it is a quadratic function.

Understanding quadratic functions helps in identifying how the function behaves, which is essential for finding their inverses. Knowing the direction the parabola opens also helps understand domain restrictions, necessary when determining the inverse.

Domain and Range

The domain of a function is the set of all possible input values (\( x \)-values) that allow the function to work without errors. The range is the set of all possible output values (\( y \)-values).

• For a quadratic function like \( f(x) = 4 - x^2 \), the domain can sometimes have restrictions. In this case, \( x \geq 0 \) to satisfy the given problem requirements. This ensures that we only consider non-negative values for \( x \).
• The range for this specific function can be determined by finding the highest and lowest points on the parabola. Since it opens downward and is centered at \( y = 4 \), the range is \( (0, 4] \).

When finding the inverse function, it's crucial to consider the range of the original function as it becomes the domain of the inverse. Our inverse function \( f^{-1}(x) = \sqrt{4 - x} \) then has a domain of \( 0 \leq x \leq 4 \) because the output of the original function, \( f(x) \), becomes the new input range.

Solving Equations

Solving equations is an essential skill used to isolate a variable or find its value in terms of other variables. It involves rearranging expressions to get the desired variable alone on one side of the equation.

• When solving for the inverse of a function, like in the exercise \( f(x) = 4 - x^2 \), we first replaced \( f(x) \) with \( y \), obtaining \( y = 4 - x^2 \).
• To isolate \( y \), we swapped \( x \) and \( y \) to form \( x = 4 - y^2 \). The task then was to solve for \( y \).
• Upon rearranging, we had \( y^2 = 4 - x \). This required taking the square root of both sides, which rendered \( y = \sqrt{4 - x} \). Importantly, we only took the non-negative root due to the domain limitation \( x \geq 0 \), ensuring that the values reflect the original function's non-negative nature.

Solutions in the world of equations are all about forming logical steps that simplify the problem, and paying attention to domain restrictions is essential for accurate inverses.