Find the inverse function $\mathrm { f } ^ { - 1 }$ of the function f.
-$f ( x ) = 3 \sin x - 2$
A) $f ^ { - 1 } ( x ) = 3 \sin ^ { - 1 } x - 2$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sin ^ { - 1 } \left( \frac { \mathrm { x } + 2 } { 3 } \right)$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \cos \left( \frac { \mathrm { x } + 2 } { 3 } \right)$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sin ^ { - 1 } \left( \frac { \mathrm { x } + 3 } { 2 } \right)$

Question

Find the inverse function $\mathrm { f } ^ { - 1 }$ of the function f.
-$f ( x ) = 3 \sin x - 2$ 
A) $f ^ { - 1 } ( x ) = 3 \sin ^ { - 1 } x - 2$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sin ^ { - 1 } \left( \frac { \mathrm { x } + 2 } { 3 } \right)$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \cos \left( \frac { \mathrm { x } + 2 } { 3 } \right)$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sin ^ { - 1 } \left( \frac { \mathrm { x } + 3 } { 2 } \right)$

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 $y = 3\sin(x) - 2$, we swap to get $x = 3\sin(y) - 2$, and then solve for $y$ to get $y = \sin^{-1}\left(\frac{x + 2}{3}\right)$.

Find the Inverse Function f−1\mathrm { f } ^ { - 1 }f−1

Find the Inverse Function $\mathrm { f } ^ { - 1 }$