functions

let f:[1/2,1]â†’[-1,1] is given by f(x)=4x³-3x then f^-1(x) is given by

1. cos (1/3 cos^-1x)
2. 3cos(cos^-1x)
3. 3sin(sin^-1x)
4. 3sin^-1(sinx)
5. sin(1/3 sin^-1x)

3 Answers

remember the formula of sin 3x or cos 3x

sir,i have already tried it x=cosA then 1/2 â‰¤ cosA â‰¤1 f(cosA)=cos3A so f^-1(cos3A)=cosA we need to find out f^-1(x) .. so x=cos3A

cos^-1x=3A

So A=1/3 (cos^-1(x))

so cos A =cos (1/3(cos^-1(x))

so f^-1(x)=cos (1/3(cos^-1(x))

well the answer matches but is the solution correct?!?!?!

yes :)

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