Paragraph: Consider the functions defined implicitly by the equation y^3-3 y+x=0 on various intervals in the real line. If x (- ,-2) (2, ) , the equation implicitly defines a unique real valued differentiable function y=f(x) . If x (-2,2) , the equation implicitly defines a unique real valued differentiable function y=g(x) , satisfying g(0)=0 . Question: _ -1 ^1 g^ (x) d x is equal to

Step-by-step solution

Let I= _ -1 ^1 g^ (x) d x=[g(x)]_ -1 ^1=g(1)-g(-1) Since, y^3-3 y+x=0 and y=g(x) (g(x))^3-3 g(x)+x=0 [by Eq. (i)] At x=1 , aligned (g(1))^3-3 g(1)+1 & =0 At x & =-1, (g(-1))^3-3 g(-1)-1 & =0 aligned On adding Eqs. (i) and (ii), we get array rlrl & & (g(1))^3+(g(-1))^3-3(g(1)+g(-1))=0 & & [g(1)+g(-1)] [(g(1))^2+(g(-1))^2-g(1) g(-1)-3 ]=0 & g(1)+g(-1)=0 & g(1) & =-g(-1) & & I & =g(1)-g(-1)=g(1)-(-g(1))=2 g(1) array