Find the inverse of the function.
-$$f ( x ) = \sqrt { 2 x + 1 }$$
A) $f ^ { - 1 } ( x ) = \frac { ( x - 1 ) ^ { 2 } } { 2 }$ for $x \geq 0$
B) $f ^ { - 1 } ( x ) = \frac { x ^ { 2 } } { 2 } - 1$ for $x \geq 0$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { 2 \mathrm { x } - 1 } { 2 }$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \frac { \mathrm { x } ^ { 2 } - 1 } { 2 }$ for $x \geq 0$

Question

Quizplus · Accepted Answer

To find the inverse, solve $y = \sqrt{2x + 1}$ for $x$: square both sides to get $y^2 = 2x + 1$, then solve for $x$ to get $x = \frac{y^2 - 1}{2}$. Thus, the inverse function is $f^{-1}(x) = \frac{x^2 - 1}{2}$ for $x \geq 0$.

Find the Inverse of the Function f(x)=2x+1f ( x ) = \sqrt { 2 x + 1 }f(x)=2x+1​

Find the Inverse of the Function $f ( x ) = \sqrt { 2 x + 1 }$