The minimum value of ( ⁻¹ x)^2 + ( ⁻¹ x)^2 is............

Step-by-step solution

Let y = ( ⁻¹ x)^2 + ( ⁻¹ x)^2 Using the identity ⁻¹ x + ⁻¹ x = 2 , we can write ⁻¹ x = 2 - ⁻¹ x Substituting this into the expression for y : y = ( ⁻¹ x)^2 + ( 2 - ⁻¹ x )^2 Let t = ⁻¹ x , where t [- 2 , 2 ] y = t^2 + ( 2 - t )^2 y = 2t^2 - t + ^2 4 y = 2 (t^2 - 2 t ) + ^2 4 y = 2 (t - 4 )^2 - 2 ( 4 )^2 + ^2 4 y = 2 (t - 4 )^2 + ^2 8 The minimum value occurs when t = 4 , which lies in the interval [- 2 , 2 ] Therefore, the minimum value is ^2 8 Answer: ^2 8