Problem 66 - A Graphical Approach to College Algebra

Question

For each function that is one-to-one, write an equation for the inverse function in the form \(y=f^{-1}(x)\) and then graph \(f\) and \(f^{-1}\) on the same axes. Give the domain and range of \(f\) and \(f^{-1}\). If the function is not one-to-one, say so. $$f(x)=-\sqrt{x^{2}-16}, x \geq 4$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \sqrt{x^2 + 16}\) for \(x \geq 0\). The domain for \(f\) is \([4, \infty)\) and range \([-\infty, -4]\); for \(f^{-1}\) domain is \([0, \infty)\) and range \([4, \infty)\).

Step 1: Verify One-to-One Property

To check whether the function \(f(x) = -\sqrt{x^2 - 16}\) is one-to-one, examine the behavior of the function over the specified domain \(x \geq 4\). A function is one-to-one if it is either strictly increasing or strictly decreasing. Since \(x \geq 4\), evaluate the derivative \(f'(x)\) to check if \(f(x)\) is monotonic:\[f(x) = -\sqrt{x^2 - 16} \Rightarrow f'(x) = \frac{-x}{\sqrt{x^2-16}}\]For \(x \geq 4\), \(x\) is always positive, hence \(-x < 0\) and the function \(f(x)\) is strictly decreasing. Therefore, \(f(x)\) is one-to-one on the given domain.

Step 2: Find the Inverse Function

To find the inverse of \(f(x)\), set \(y = -\sqrt{x^2 - 16}\) and solve for \(x\) in terms of \(y\). Start by squaring both sides:\[y^2 = x^2 - 16\]Rearrange to solve for \(x^2\):\[x^2 = y^2 + 16\]Solve for \(x\) according to the domain restriction \(x \ge 4\):\[x = \sqrt{y^2 + 16}\]Thus, the inverse function is \(f^{-1}(y) = \sqrt{y^2 + 16}\) for \(y \leq 0\). Thus in terms of \(x\):\[f^{-1}(x) = \sqrt{x^2 + 16}\]

Step 3: Graph the Functions

Graph the original function \(f(x) = -\sqrt{x^2 - 16}\) for \(x \geq 4\) as a decreasing curve starting at \((4, -\sqrt{16}) = (4, -4)\) and moving leftward from there. Alternatively, plot the inverse function \(f^{-1}(x) = \sqrt{x^2 + 16}\), but since it requires \(y \le 0\), restrict the graph to valid \(x\)-values.The graph of \(f^{-1}(x)\) starts at the point \((0, 4)\) and moves upward. To visualize the connection, note that these two graphs are reflections of one another over the line \(y = x\).

Step 4: Determine Domain and Range

For the function \(f(x) = -\sqrt{x^2 - 16}\), the domain is \(x \geq 4\) (since \(x^2 - 16\) must be non-negative), and the range is \([-\infty, -4]\).For the inverse function \(f^{-1}(x) = \sqrt{x^2 + 16}\), the domain is \([0, \infty)\) because \(y\) is computed based on \(y^2\), and the range is \([4, \infty)\).

Key Concepts

One-to-One Functions

A function is called "one-to-one" when each input has exactly one unique output, and vice versa. This means that no horizontal line intersects the graph of the function more than once. This property is essential for a function to have an inverse that is also a function. In simpler terms, a one-to-one function does not repeat any output value.
To determine if a function is one-to-one, we often check if it is consistently increasing or decreasing, which is known as being monotonic. For the function in the exercise, \[f(x) = -\sqrt{x^2 - 16}\]we examined if it is consistently behaving in one direction.
We derived\[f'(x) = \frac{-x}{\sqrt{x^2-16}}\]which shows that for \(x \geq 4\), \(f(x)\) is decreasing because the derivative is negative. So, indeed \(f(x)\) is one-to-one over its given domain. Recognizing this property allows us to find an inverse function, ensuring every output corresponds uniquely to an input.

Domain and Range

Understanding the concepts of Domain and Range is crucial when dealing with functions and their inverses.

• Domain: The set of all possible input values for the function. It's like the limits within which the function operates.
• Range: The set of all possible output values. It tells you the limits of the values you can expect to get from a function.

For the function \(f(x) = -\sqrt{x^2 - 16}\), the domain is \(x \geq 4\) because \(x^2 - 16\) must be non-negative, making sure the expression under the square root is valid. The range of this function is \([-fty, -4]\) as it produces negative values starting from \(-4\) and going downwards.
When we find the inverse \(f^{-1}(x) = \sqrt{x^2 + 16}\), the domain switches to \([0, \infty)\) since we are dealing with a one-to-one function reflected in terms of output. The range is \([4, \infty)\), showcasing how these properties switch between a one-to-one function and its inverse.

Graphing Functions

Graphing both a function and its inverse on the same set of axes can illustrate their relationship vividly. When you plot a function and its inverse, the graphs are mirror images across the line \(y = x\). This line acts as the mirror, reflecting how each point on one graph corresponds to a point on the inverse graph with the coordinates swapped.
For \(f(x) = -\sqrt{x^2 - 16}\), only the part where \(x \geq 4\) is plotted because of its domain. The graph decreases from the point \((4, -4)\), reflecting its strictly decreasing nature. Meanwhile, the inverse function \(f^{-1}(x) = \sqrt{x^2 + 16}\) is plotted starting from the equivalent turnaround point \((0, 4)\), heading upward with \(x \geq 0\).
When graphing, always consider the specified domains, ensuring that each function stays within its operational boundaries. This visual method helps confirm the calculations and intuitively understand how the function transitions into its inverse form.