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Find the greatest and least values of the function `f(x)=(sin^(-1)x)^3+(cos^(-1)x)^3,-1 le x le 1`

Answer» Using `sin^(-1)x+cos^(-1)x=pi/2` we can write the function in terms of `sin^(-1)x(or cos^(-1)x)` alone. And then note that range of `sin^(-1)x " is "[-pi/2,pi/2] and cos^(-1)x " is "[0,pi]`.
Put `cos^(-1)x=t, ` so that `sin^(-1)x=pi/2-t, and 0, let le pi`
Now the trigonometry ends and algebra takes over. Finding the greatest and least values of the function f(x) amounts to finding the greatest and least values of function g(t).
where
`g(t)=(pi/2-t)^3+t^3,0 le t le pi`
`=(pi/2-t+t^3)-3(pi/2-t)(t)(pi/2-t+t)`
`=(pi/2)^3-3t(pi/2-t)pi/2`
`pi^3/8-3pi^3/4t+(3pi)/2t^2`
`(3pi)/2[t^2-pi/2t+pi^2/12]`
`=(3pi)/2[(t0pi/4)^2+pi^2/12-pi^2/16]`
`=(3pi)/2[(t-pi/2)^2+pi^2/48]`
`g(pi/4)=(3pi)/2.pi^2/48=pi^3/32`
`g(0)=pi^3/8`
`g(pi)=(3pi)/2[((3pi)/4)^2+pi^2/48]`
`=(3pi)/2[(9pi^2)/16+pi^2/48]`
`=(3pi)/2.(28pi^2)/48`
`=7/8pi^3`
So the least value is `pi^3/32`, attained at `t=pi/4`,
i.e., `x=1/sqrt2`
and the greatest value is `(7pi^3)/8`, attained at `t=pi,i.e., x= -1`


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