Question

Let the function f defined by f(x)
= 2x+1/1- 3x'
then f^-1 (x) is​

Answer 1

Answer:

f(x)=2

x(x+1)

, i.e., y=2

x(x+1)

,

⇒log

2

y=x(x−1)=x

2

−x

⇒x

2

−x−log

2

y=0

⇒x=

2

1±

1+4log

2

y

Since x∈[1,∞) ∴ -ve sign is ruled out.

∴⇒x=

2

1

(1+

1+4log

2

y

)

⇒f

−1

(x)=

2

1

(1+

1+4log

2

y

)

⇒f

−1

(x)=

2

1

(1+

1+4log

2

x

)