a fxn is defined on[-1,1]and area of sq. is 4. with its vertices at (0,0) and (x,f(x)) then f(x) is...

Dear Jauneet

one vertex of square is given (0,0) .then a square of area four can not be formed whose rest three  point  lie in [-1,1]

so such function is not possible,

if (x,f(x) is the vertix of other side of diagonal,and other two vertix can be outside the [-1,1] then square is possible

then length of diagonnal = √{f²(x) +x²}

so area =4 =(digonal)²/2

     4= {f²(x) +x²}/2

 so f(x) =±√(8-x²)

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