Question

f(x)=log(x+√(x²+1)) How to find the values f^-1(x) and f^-1(y)???...

Please answer this question guys it's urgent........​

Answer 1

Let y=log(x+(x2+1)−−−−−−−√ )

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√ (ey−x)2=x2+1

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√ (ey−x)2=x2+1 e2y−2xey+x2=x2+1

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√ (ey−x)2=x2+1 e2y−2xey+x2=x2+1 e2y−1=2xey

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√ (ey−x)2=x2+1 e2y−2xey+x2=x2+1 e2y−1=2xey x=e2y−12eyex

Let y=log(x+(x2+1)−−−−−−−√ )So. ey=x+(x2+1)−−−−−−−√ ie. ey−x=(x2+1)−−−−−−−√ (ey−x)2=x2+1 e2y−2xey+x2=x2+1 e2y−1=2xey x=e2y−12eyex So. f−1(x)=12(e(2x)−1)

Step-by-step explanation:

please mark my Brainliest