91Ó°ÊÓ

For each piece of the given function, we will find the inverse by swapping \(x\) and \(y\) (where \(y = f(x)\)), and then solving for \(y\). 1. For the first piece \(2x - 1\), if \(x < 1\): Swap \(x\) and \(y\): \(x = 2y - 1\) Solve for \(y\): \(y = \frac{1}{2}x + \frac{1}{2}\) 2. For the second piece \(\sqrt{x}\), if \(1 \leq x < 4\): Swap \(x\) and \(y\): \(x = \sqrt{y}\) Solve for \(y\): \(y = x^2\) 3. For the third piece \(\frac{1}{2}x^2 - 6\), if \(x \geq 4\): Swap \(x\) and \(y\): \(x = \frac{1}{2}y^2 - 6\) Solve for \(y\): \(y = \pm\sqrt{2(x+6)}\) (note that we'll consider the correct sign later when defining domain)

The domain for the inverse function is the range of the original function. So, let's find the boundaries: 1. For the first piece, \(x < 1\), therefore: \(y = 2(1) - 1 = 1\) This means the range of the first piece goes from negative infinity to 1 (not inclusive). So, the domain for the first piece of the inverse function is \(-\infty < x < 1\). Using the domain and inverse function we found before, the inverse piece is: \(y = \frac{1}{2}x + \frac{1}{2}\), if \(-\infty < x < 1\). 2. For the second piece, \(1 \leq x < 4\), therefore: \(y = \sqrt{1} = 1\) \(y = \sqrt{4} = 2\) This means the range of the second piece goes from 1 (inclusive) to 2 (not inclusive). So, the domain for the second piece of the inverse function is \(1 \leq x < 2\). Using the domain and inverse function we found before, the inverse piece is: \(y = x^2\), if \(1 \leq x < 2\). 3. For the third piece, \(x \geq 4\), therefore: \(y = \frac{1}{2}(4)^2 - 6 = 2\) This means the range of the third piece starts from 2 (inclusive) and goes to positive infinity. So, the domain for the third piece of the inverse function is \(2 \leq x < \infty\). We also need to consider the correct sign for the inverse function (we had two solutions with different signs). Since the range starts at 2, we should use the positive square root: Using the domain and inverse function we found before, the inverse piece is: \(y = \sqrt{2(x+6)}\), if \(2 \leq x < \infty\).

Now, we combine all the inverse pieces into one piecewise inverse function: $$ f^{-1}(x)=\left\\{\begin{array}{ll} \frac{1}{2}x + \frac{1}{2} & \text { if } -\infty < x < 1 \\\ x^{2} & \text { if } 1 \leq x < 2 \\\ \sqrt{2(x+6)} & \text { if } x \geq 2 \end{array}\right. $$ The domain of \(f^{-1}(x)\) is \(-\infty < x < \infty\).