The function $f ( x ) = \sqrt { x ^ { 2 } - 5 }$ is one-to-one on the domain $x \leq - \sqrt { 5 }$. Find $f ^ { - 1 } ( x )$.
A) $\quad f ^ { - 1 } ( x ) = x ^ { 2 } + 5$
B) $f ^ { - 1 } ( x ) = \sqrt { x ^ { 2 } + 5 }$
C) $f ^ { - 1 } ( x ) = \frac { 1 } { \sqrt { x ^ { 2 } - 5 } }$
D) $\quad f ^ { - 1 } ( x ) = - \sqrt { x ^ { 2 } + 5 }$
E) $f ^ { - 1 } ( x ) = - \sqrt { x ^ { 2 } - 5 }$

Question

Quizplus · Accepted Answer

To find the inverse function $f^{-1}(x)$, we swap $x$ and $y$ in the original function and solve for $y$. Starting with $y = \sqrt{x^2 - 5}$, we swap to get $x = \sqrt{y^2 - 5}$, then square both sides to eliminate the square root, resulting in $x^2 = y^2 - 5$. Solving for $y$, we get $y^2 = x^2 + 5$, and taking the square root gives $y = \pm\sqrt{x^2 + 5}$. Since the domain is $x \leq -\sqrt{5}$, the function is decreasing, and we choose the negative branch to maintain the function's behavior, resulting in $f^{-1}(x) = -\sqrt{x^2 + 5}$.

The Function f(x)=x2−5f ( x ) = \sqrt { x ^ { 2 } - 5 }f(x)=x2−5​

The Function $f ( x ) = \sqrt { x ^ { 2 } - 5 }$