Problem 83 - Precalculus Essentials

Question

Use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one). $$f(x)=-\sqrt{16-x^{2}}$$

Step-by-Step Solution

Answer

The function \(f(x)=-\sqrt{16-x^{2}}\) has an inverse that is also a function. That means the function is one-to-one.

Step 1: Plotting the Function

Use a graphing utility to plot the function \(f(x)=-\sqrt{16-x^{2}}\). Because square-rooted numbers cannot be negative, the domain of this function is \(-4 \leq x \leq 4\). This function represents a semicircle with radius 4, centered at the origin, and lying below the x-axis.

Step 2: Apply the Horizontal Line Test

The horizontal line test is used for determining whether the function is one-to-one. If a horizontal line intersects the graph of a function at more than one point, then the function is not one-to-one. After drawing the graph, apply the horizontal line test to it. In this case, any horizontal line will intersect the graph at most one point.

Step 3: Conclusion

Since any horizontal line intersects the graph at most one point, it can be concluded that the function is one-to-one. Therefore, the function has an inverse that is also a function.

Key Concepts

Horizontal Line Test

The horizontal line test is an essential tool for checking if a function is one-to-one. A one-to-one function corresponds to having exactly one unique output for every input.The significance of a function being one-to-one is that it can have an inverse which is also a function.
To use the horizontal line test, you need to draw a horizontal line across the graph of the function. If any horizontal line crosses the graph at more than one point, the function is not one-to-one.

• For example, let’s examine the function \(f(x)=-\sqrt{16-x^{2}}\) which forms a semicircle.
• If you draw a horizontal line across this semicircle, it will intersect at only one point, meaning each y-value comes from only one x-value.
• Therefore, it passes the horizontal line test, confirming it is one-to-one.

This result permits us to say the function has an inverse that is also a function.

Function Inverse

A function inverse essentially "undoes" the function, taking an output back to its input. To find a function's inverse, you swap the roles of the dependent and independent variables and solve for the new variable. If \(y = f(x)\), then the inverse function, usually denoted as \(f^{-1}(x)\), will satisfy \(x = f^{-1}(y)\).
In the context of \(f(x)=-\sqrt{16-x^{2}}\), knowing it is one-to-one, confirms an inverse exists. When a function is not one-to-one, any given output would correspond to multiple inputs, which makes it impossible to uniquely trace back to the original input.

• This is crucial because, without a one-to-one relationship, the inverse cannot itself be a function.
• Only one output should correlate back to exactly one input to maintain functionality.

Understanding inverse functions broaden your grasp of how inputs and outputs correlate in different settings.

Domain and Range

Domain and range are foundational concepts when working with functions.
The domain of a function refers to all possible input values (x-values), while the range consists of all possible output values (y-values). For \(f(x)=-\sqrt{16-x^{2}}\), the domain is restricted to \(-4 \leq x \leq 4\). This limitation arises because the expression inside the square root, \(16-x^{2}\), must be non-negative, thereby limiting x-values to those within the semicircle.
Similarly, the range for \(f(x)\) is \(-4 \leq f(x) \leq 0\), because the function covers the negative y-values in the semicircle below the x-axis.

• Recognizing these constraints helps define where the function is applicable.
• The domain and range are crucial in deducing both the behavior of the function and its inverse.

The domain of the inverse is the range of the original function, and vice versa. This interrelation is fundamental when evaluating the applicability of functions in mathematical and real world problems.