Problem 36 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 25-42: Find the inverse function of \(f\). $$ f(x)=\sqrt{4-x^{2}}, 0 \leq x \leq 2 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(y) = \sqrt{4 - y^{2}} \) for \( 0 \leq y \leq 2 \).

Step 1: Understand the function

The function given is \( f(x) = \sqrt{4-x^{2}} \) and it is defined for \( 0 \leq x \leq 2 \). This is a semi-circle above the x-axis with a radius of 2, centered at the origin.

Step 2: Set up the equation for the inverse

To find the inverse function, we need to express \( x \) in terms of \( y \) by setting \( y = f(x) \). So, \( y = \sqrt{4 - x^{2}} \).

Step 3: Solve for \( x \) in terms of \( y \)

Square both sides to eliminate the square root: \( y^{2} = 4 - x^{2} \). Rearrange this to \( x^{2} = 4 - y^{2} \). Then, solve for \( x \): \( x = \sqrt{4 - y^{2}} \).

Step 4: Determine the correct branch of the inverse function

Given the constraint \( 0 \leq x \leq 2 \), the principal square root should be considered. Therefore the inverse function in terms of \( y \) should be: \( f^{-1}(y) = \sqrt{4 - y^{2}} \).

Step 5: Specify the domain of the inverse function

Since \( f(x) \) is defined for \( 0 \leq x \leq 2 \), its corresponding range is \( 0 \leq y \leq 2 \). Thus, the inverse function \( f^{-1}(y) = \sqrt{4 - y^{2}} \) is defined for \( 0 \leq y \leq 2 \).

Key Concepts

Semi-Circle Function

In mathematics, a semi-circle function is often associated with functions that have a square root form involving a squared term. In the exercise above, we're dealing with the function \( f(x) = \sqrt{4-x^2} \). If you think of a circle, it has the equation \( x^2 + y^2 = r^2 \), where \( r \) is the radius. For a semi-circle, you're effectively taking half of this figure, which is why it is called a semi-circle. This semi-circle is placed above or below the x-axis depending on the other parts of the equation.
This function is defined for \( 0 \leq x \leq 2 \), meaning it's the upper part of the circle with radius 2, centered at the origin. Visualizing this can help you grasp what the function is doing: it's like having half of a pizza outline. Only the crust from \( x=0 \) to \( x=2 \) is what the function encompasses. The new points on this arc represent the possible values of \( f(x) \).

• Upper semi-circle nature because the square root is non-negative
• Radius here is 2 (indicated by \( \sqrt{4} \))
• Helps to understand the properties of specific function types

Domain and Range

When dealing with functions, two crucial concepts are domain and range. The domain of a function is all the possible input values (\( x \)-values) that the function can accept without causing any errors, like dividing by zero or taking the square root of a negative number. Meanwhile, the range is all the potential outcomes (\( y \)-values) the function can output.
For our semi-circle function, \( f(x) = \sqrt{4-x^2} \), the domain is \( 0 \leq x \leq 2 \). Why? Because if \( x \) goes beyond these limits, the expression under the square root becomes negative, which is not possible for real numbers in this context.
The range, or the possible outputs, is \( 0 \leq y \leq 2 \). This ensures that the values coming out of the function are the actual y-values seen when you plot the semi-circle.

• Domain: What goes into the function
• Range: What comes out of the function
• Important for defining both functions and their inverses

Principal Square Root

The concept of the principal square root is key when dealing with functions involving square roots. The principal square root is essentially the non-negative square root of a number. Instead of possibly having two roots, one positive and one negative, we choose only the positive one.
For example, in our function \( f(x) = \sqrt{4-x^2} \), the square root ensures the value is always positive or zero. As a square root operation inherently implies there could be both a positive and negative solution, the principal square root simplifies this by focusing only on results that are non-negative, which fits perfectly for functions that map to a real-world context.
The principal square root is why, in the solution steps, we directly write \( x = \sqrt{4 - y^2} \) for determining the inverse, eliminating any ambiguity about what value of \( x \) we are referring to under the square root context.

• Chooses non-negative root
• Avoids complex or ambiguous solutions
• Applies more easily to real-world problems