Let `f:R to R` and `h:R to R` be differentiable functions such that `f(x)=x^(3)+3x+2,g(f(x))=x and h(g(g(x)))=x` for all `x in R`. Then, find the value of h'(1).

Answer

`h'(1)=666`

Explanation

`f(x)=x^(3)+3c+2`
`therefore" "f(1)=6`
`therefore" In " g(f(x))=x," Putting "x=1`
`g(6)=1`
`"Also "g(f(x))=x`
`rArr" "g'(f(x))xxf'(x)=1`
`"Put "x=0`
`therefore" "g'(f(0))cdotf'(0)=1`
`therefore" "g'(2)=(1)/(f'(0))=(1)/(3)`
`f(3)=38`
`therefore" "g(38)=3`
`therefore" "h(g(3))=h(g(g(38)))=38`
`f(2)=16 rArr g(16)=2`
`therefore" "h(g(g(16))=h(g(2))=h(0)`
`therefore" "16=h(g(g(16))=h(0)`
`therefore" Option (3) is correct."`
`f'(x)=3x^(2)+3`
`therefore" "f'(6)=111, f'(1)=6rArrg'(6)=(1)/(6)`
`h(g(g(x)))=x`
`rArr" "h'(g(g(x)))xxg'(g(x))xxg'(x)=1`
`"Put "x=236,`
`therefore" "h'(g(g(236)))xxg'(g(236))xxg'(236)=1`
`rArr" "h'(g(6))g'(6)xx(1)/(g'(6))=1`
`rArr" "h'(1)=666`