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If `'f'` is an increasing function from `RvecR` such that `f^(x)>0a n df^(-1)` exists then `(d^2(f^(-1)(x)))/(dx^2)` is `<0` b. `>0` c. `=0` d. cannot be determined

A

lt 0

B

gt 0

C

0

D

cannot be determined

Text Solution

Verified by Experts

The correct Answer is:
A
We have `f'(x)gt0 and f''(x)gt0`
Let `g(x)=f^(-1)(x)`
`f(g(x))=x`
`f'(g(x)).g'(x)=1`
`g'(x)=(1)/(f'(g(x)))`
`therefore" "g'(x)gt0`
`g''(x)=-(1)/((f'(g(x)))^(2))f''(g(x))g'(x)`
`lt0" "(because f''(x)lt0)`
`therefore" "(d^(2)(f^(-1)(x)))/(dx^(2))=g''(x)lt0`
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