The function f (x)= [ (x+ 1+x^2 ) ] is _______ function

Step-by-step solution

To determine the parity of f(x)= [ (x+ 1+x² ) ] , we evaluate f(-x) : f(-x) = [ (-x+ 1+(-x)² ) ] = [ (-x+ 1+x² ) ] Observe that the product (x+ 1+x² ) (-x+ 1+x² ) = (1+x²) - x² = 1 , which implies: -x+ 1+x² = 1 x+ 1+x² Substituting back: f(-x) = [ ( 1 x+ 1+x² ) ] Applying the logarithmic identity (1/A) = - A : f(-x) = [- (x+ 1+x² ) ] Since secant is an even function ( (- ) = ) , we obtain: f(-x) = [ (x+ 1+x² ) ] = f(x) This equality f(-x) = f(x) confirms that the function is even. A