Question

If the function f : [1,00 ) → [1,00) is
defined by f(x) = 2^x(x-1) then f^-1(x) is

Answer 1

Answer:

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Answer 2

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

)

Step-by-step explanation:

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