find Inverse function. [ f(x)=sqrt(x^2)-5 ]

To find the inverse function of \( f(x) = \sqrt{x^2 - 5} \), we need to follow these steps:<br /><br />1. **Express \( y \) in terms of \( x \):**<br /> \[<br /> y = \sqrt{x^2 - 5}<br /> \]<br /><br />2. **Swap \( x \) and \( y \):**<br /> \[<br /> x = \sqrt{y^2 - 5}<br /> \]<br /><br />3. **Solve for \( y \):**<br /><br /> - First, square both sides to eliminate the square root:<br /> \[<br /> x^2 = y^2 - 5<br /> \]<br /><br /> - Then, solve for \( y^2 \):<br /> \[<br /> y^2 = x^2 + 5<br /> \]<br /><br /> - Finally, take the square root of both sides:<br /> \[<br /> y = \pm \sqrt{x^2 + 5}<br /> \]<br /><br />4. **Determine the domain and range:**<br /><br /> - The original function \( f(x) = \sqrt{x^2 - 5} \) is defined for \( x^2 - 5 \geq 0 \), which means \( x \leq -\sqrt{5} \) or \( x \geq \sqrt{5} \).<br /> - The range of \( f(x) \) is \( [0, \infty) \) because the square root function outputs non-negative values.<br /><br /> Since the inverse function must be a function (i.e., it must pass the vertical line test), we restrict the domain of the original function to \( x \geq \sqrt{5} \). This ensures that the original function is one-to-one and thus has an inverse.<br /><br />5. **Write the inverse function:**<br /><br /> Given the restriction \( x \geq \sqrt{5} \), the inverse function is:<br /> \[<br /> f^{-1}(x) = \sqrt{x^2 + 5}<br /> \]<br /><br />So, the inverse function of \( f(x) = \sqrt{x^2 - 5} \) is:<br />\[<br />f^{-1}(x) = \sqrt{x^2 + 5}<br />\]