f(x) = $\frac { e ^ { x } + 1 } { e ^ { x } - 1 }$ Enter your answer exactly as just $\frac { \mathrm { P } \left( \mathrm { e } ^ { \mathrm { x } } \right) } { \left( \mathrm { Q } \left( \mathrm { e } ^ { \mathrm { x } } \right) \right) ^ { \mathrm { n } } }$ where P and Q are polynomials in $e ^ { x }$ in standard form.

Question

Quizplus · Accepted Answer

The Answer of f(x) = $\frac { e ^ { x } + 1 } { e ^ { x } - 1 }$ Enter your answer exactly as just $\frac { \mathrm { P } \left( \mathrm { e } ^ { \mathrm { x } } \right) } { \left( \mathrm { Q } \left( \mathrm { e } ^ { \mathrm { x } } \right) \right) ^ { \mathrm { n } } }$ where P and Q are polynomials in $e ^ { x }$ in standard form.

F(x) = ex+1ex−1\frac { e ^ { x } + 1 } { e ^ { x } - 1 }ex−1ex+1​

F(x) = $\frac { e ^ { x } + 1 } { e ^ { x } - 1 }$