f(x)=2 x- ⁻¹ x- (x+ x^2+1 ) is monotonically increasing, when

Step-by-step solution

The function is given by f(x) = 2x - ⁻¹ x - (x + x^2 + 1 ) . To determine the interval where f(x) is monotonically increasing, we find its derivative f'(x) with respect to x . f'(x) = d dx (2x) - d dx ( ⁻¹ x) - d dx ( (x + x^2 + 1 )) . The derivative of ⁻¹ x is 1 1 + x^2 . The derivative of (x + x^2 + 1 ) is 1 x + x^2 + 1 (1 + 2x 2 x^2 + 1 ) = 1 x + x^2 + 1 ( x^2 + 1 + x x^2 + 1 ) = 1 x^2 + 1 . Thus, f'(x) = 2 - 1 1 + x^2 - 1 x^2 + 1 . Let t = 1 x^2 + 1 . Since x^2 0 , we have x^2 + 1 1 , so 0 Then 1 1 + x^2 = t^2