Let f(x) = cases ⁻¹( ^2 - 3) - x, & x Then the set of all values of , for which f(x) has a local minimum at x=0 , is:

Step-by-step solution

Given, f(x) = cases ⁻¹( ^2 - 3) - x, & x Now f(0) = 6 For x f(x) is decreasing. For x > 0 , f'(x) = 3x^2 + 2 > 0 f(x) is increasing. So, at x=0 there is a possibility of a point of local minima. For a local minimum at x=0 , we must have f(0) _ x 0^- f(x) 6 ⁻¹( ^2 - 3) Since ⁻¹(y) is an increasing function, we have: ^2 - 3 ( 6 ) = 1 2 ^2 7 2 Also, for the domain of ⁻¹( ^2 - 3) , we must have: -1 ^2 - 3 1 2 ^2 4 Taking the intersection of the conditions, we get: 7 2 ^2 4 [-2, - 7 2 ] [ 7 2 , 2 ] Answer: [-2, - 7 2 ]