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let f;(-1,1)-R be a differentiable function with f(0)=-1 ,f|(o)=1,g(x)=[f(2f(x)+2)] square ,g(0)= 
Asked by mishrk261 | 28 Aug, 2019, 21:20: PM
answered-by-expert Expert Answer
g(0)=[f(2f(0)+2)]2
=[f(2 x (-1)+2)]2
=[f(-2+2)]2
=[f(0)]2
=[-1]2
=1
Hence, g(0)=1
Answered by Renu Varma | 29 Aug, 2019, 11:35: AM

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