91Ó°ÊÓ

When assessing a function, understanding its domain and range is crucial. The **domain** refers to all possible input values, while the **range** signifies all possible outputs.
For the function \( f(t) = \sqrt{1-t^2} \), the domain comprises values that keep the expression inside the square root non-negative, meaning \( 1 - t^2 \geq 0 \). Solving this inequality, we find that the domain is \(-1 \leq t \leq 1\).

• The domain is influenced by ensuring \( t^2 \leq 1 \), covering the range from -1 to 1.
• The range is determined by the values that \( \sqrt{1-t^2} \) can produce, confined to \([0, 1]\).

Knowing the domain and range is essential, especially when evaluating the invertibility of a function. When a function has an inverse, the domain of the original function becomes the range of the inverse, and vice versa.
The process of finding domain and range sets up the foundation for working with any inverse functions.

For a function to qualify as having an inverse, it must be **bijective**--a function that is both **injective** and **surjective**. This ensures each output is uniquely paired with an input, making it one-to-one and onto.
The function \( f(t) = \sqrt{1-t^2} \) isn't injective over its initial domain, as different input values may yield the same result. For instance, \( f(-\frac{1}{2}) \) and \( f(\frac{1}{2}) \) both equal \( \frac{\sqrt{3}}{2} \). Therefore, it fails the injectivity test across \([-1, 1]\).

• **Injective (One-to-One):** A function is injective if every output corresponds to one unique input.
• **Surjective (Onto):** A function is surjective if every possible output is achieved by some input.

To make a non-bijective function bijective, we can **restrict its domain**. For this particular function, restricting the domain to \([0,1]\) resolves the injective issue, making the function strictly decreasing over this interval. The bijective adjustment enables the existence of an inverse.

Graphing a function and its inverse visually illustrates their relationship, especially through the concept of symmetry about the line \( y = x \).
The given function \( f(t) = \sqrt{1-t^2} \) represents the upper half of a circle centered at the origin with radius 1, strictly over the domain \([0, 1]\). Thus, its graph forms a semicircle from (0,1) to (1,0).

• The graph of the original function gives a visual understanding of its output range for the input values.
• The inverse \( f^{-1}(y) = \sqrt{1-y^2} \) mirrors this semicircle along the line \( y = x \).

Reflecting the graph over the line \( y = x \) ensures this mirror is evident, showcasing their symmetry--a hallmark of inverse functions. This visual guide solidifies conceptual understanding, as students can see how inputs and outputs swap places in inverse functions.