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Chapter 2: Problem 41

a. Find an equation for \(f^{-1}(x)\) b. Graph \(f\) and \(f^{-1}\) in the same rectangular coordinate system. c. Use interval notation to give the domain and the range of \(f\) and \(f^{-1}\) $$f(x)=x^{2}-4, x \geq 0$$

Short Answer

Expert verified
The inverse of the function \(f(x) = x^2 - 4, x \geq 0\) is \(f^{-1}(x) = \sqrt{x + 4}, x \geq -4\). The domain and range of \(f(x)\) are \([0, \infty)\) and \([-4, \infty)\) respectively, while those of \(f^{-1}(x)\) are \([-4, \infty)\) and \([0, \infty)\) respectively.

Step by step solution

01

Finding the Inverse

In order to find the inverse \(f^{-1}(x)\) of a function \(f(x)\), it is necessary to replace \(f(x)\) with \(y\), interchange \(x\) and \(y\) and then solve for \(y\). Replacing \(f(x)\) with \(y\), we get \(y = x^2 - 4\). Switching \(x\) and \(y\), we then get \(x = y^2 - 4\). Solving this equation for \(y\), we have \(y = \sqrt{x + 4}\). Since the original function \(f(x)\) was defined for \(x \geq 0\), the inverse function \(f^{-1}(x) = \sqrt{x + 4}\) is defined for \(x \geq -4\). So \(f^{-1}(x) = \sqrt{x + 4}, x \geq -4\).
02

Graphing the Function and Its Inverse

The graph of the initial function \(f(x) = x^2 - 4\) is a parabola, opening upwards with the vertex at \((0, -4)\) shifted down 4 units from the graph of \(y = x^2\). The graph of its inverse \(f^{-1}(x) = \sqrt{x+4}\) is a half parabola that opens to the right. The point of symmetry between the original function and its inverse is located at the origin (\(0, 0)\).
03

Determining the Domain and Range

The domain of \(f(x) = x^2 - 4\) can be set as \(x \geq 0\) whereas the range is restricted to all real numbers that are greater than or equal to -4, i.e., \(y \geq -4\). On the other hand, the domain of \(f^{-1}(x) = \sqrt{x+4}\) is \(x \geq -4\) because it cannot take any values less than -4, while its range will be all non-negative real numbers, i.e., \(y \geq 0\).

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