Find Inverse of Square Root Functions

• Domain of \(f\): \([1, +\infty)\), Range of \(f\): \([0, +\infty)\).
• Write \(f\) as an equation: \( y = \sqrt{x - 1} \)
• Square both sides: \( y^2 = x - 1 \)
• Solve for \(x\): \( x = y^2 + 1 \)
• Swap \(x\) and \(y\) to get the inverse: \( f^{-1}(x) = x^2 + 1 \)

Conclusion: Domain and range of \(f^{-1}\): Domain: \([0, +\infty)\), Range: \([1, +\infty)\).

• Domain: \(x + 3 \ge 0 \Rightarrow x \ge -3\) → \([-3, +\infty)\)
• Range: \([-5, +\infty)\)
• Write as equation: \( y = \sqrt{x + 3} - 5 \Rightarrow \sqrt{x + 3} = y + 5 \)
• Square both sides: \( x + 3 = (y + 5)^2 \)
• Solve for \(x\): \( x = (y + 5)^2 - 3 \)
• Swap \(x\) and \(y\): \( f^{-1}(x) = (x + 5)^2 - 3 \)

Conclusion: Domain and range of \(f^{-1}\): Domain: \([-5, +\infty)\), Range: \([-3, +\infty)\).

• Domain: \((-\infty, -1]\), Range: \((-\infty, 0]\)
• Write as equation: \( y = -\sqrt{x^2 - 1} \)
• Square both sides: \( y^2 = x^2 - 1 \Rightarrow x^2 = y^2 + 1 \)
• Select solution using domain (\(x \le -1\)): \( x = -\sqrt{y^2 + 1} \)
• Swap \(x\) and \(y\) for the inverse: \( f^{-1}(x) = -\sqrt{x^2 + 1} \)

Conclusion: Domain and range of \(f^{-1}\): Domain: \((-\infty, 0]\), Range: \((-\infty, -1]\).