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In Exercises \(49-62,\) (a) find the inverse function of \(f\) (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\) , and (d) state the domain and range of \(f\) and \(f^{-1}\) . $$ f(x)=\sqrt{4-x^{2}}, \quad 0 \leq x \leq 2 $$

Short Answer

Expert verified
The inverse function is \(f^{-1}(x) = \sqrt{4 - x^2}\). Both \(f(x)\) and \(f^{-1}(x)\) are plotted as semi-circles. They reflect about the line \(y = x\). The domain and range for both the original function and its inverse are \(0 \leq x \leq 2\) and \(0 \leq y \leq 2\), respectively.

Step by step solution

01

Find the Inverse Function

We start by setting \(y = f(x)\), yielding the equation \(y = \sqrt{4-x^2}\). To find the inverse, we switch \(y\) and \(x\), then solve it for \(y\). This gives us \(x = \sqrt{4 - y^2}\). Solving for \(y\) gives us \(y = \sqrt{4 - x^2}\) and \(y = -\sqrt{4 - x^2}\). Since the given domain is \(0 \leq x \leq 2\), the inverse function becomes \(f^{-1}(x) = \sqrt{4 - x^2}\).
02

Graph

Next, we graph both functions. \(f(x)\) is a semi-circle above the \(x\)-axis with radius 2, while \(f^{-1}(x)\) is a semi-circle to the right of the \(y\)-axis with the same radius. These graphs are mirror images over the line \(y=x\).
03

Describe the Relationship

The function \(f(x)\) and its inverse \(f^{-1}(x)\) are reflections of each other across the line \(y = x\). When a function has an inverse, this is always the relationship between their graphs.
04

Determine the Domain and Range

The original function \(f(x)\) is defined for \(0 \leq x \leq 2\), with range \(0 \leq y \leq 2\). For the inverse function \(f^{-1}(x)\), the domain and range reverse, making the domain \(0 \leq x \leq 2\) and the range \(0 \leq y \leq 2\).

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Most popular questions from this chapter

SPORTS The winning times (in minutes) in the women's 400-meter freestyle swimming event in the Olympics from 1948 through 2008 are given by the following ordered pairs. \((1948, 5.30)\) \((1952, 5.20)\) \((1956, 4.91)\) \((1960, 4.84)\) \((1964, 4.72)\) \((1968, 4.53)\) \((1972, 4.32)\) \((1976, 4.16)\) \((1980, 4.15)\) \((1984, 4.12)\) \((1988, 4.06)\) \((1992, 4.12)\) \((1996, 4.12)\) \((2000, 4.10)\) \((2004, 4.09)\) \((2008, 4.05)\) A linear model that approximates the data is \(y = -0.020t + 5.00\), where \(y\) represents the winning time (in minutes) and \(t=0\) represents 1950. Plot the actual data and the model on the same set of coordinate axes. How closely does the model represent the data? Does it appear that another type of model may be a better fit? Explain. (Source: International Olympic Committee)

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