Could someone please check my work

2. Find f'1([3, 10]):: on u [132} L e T 103 Iii9 1R_,_,._{',?CEEF3 5'1 1%fo (m 33'\" (- Explain: f\"(i3,101)={x 6 szlm 6 {3,101} = i~ 6, ~ 5] U {1,21 by notation 2.3.13 for the preimage of [3, 10] E R . Taking the inverse of f(:c) means a; = (f \"1(m) + 2)2 - 6 , which means a: + 6 : (f "'1(a:) + 2)2 .Squaring both sides [1/31 + 3] = f*1(:c) + 2 , which means f\"1(a:) = Na: + 6| - 2 . Then substituting x=3 into f'1(x) means f1(3) = [V3 + 6' 2 , which means f_1(3) = [3| - 2 , which means f_1(3) = { - 5, 1} . Now substituting x=10 into f'1(x) means ; f~1(10) = [V 10 + 6| 2 , which means f_1(10) = [4| 2 , which means f1(10)={~6,}. i The graph of f is a parabolic function. Since the graph of f has a positive coefficient of x, f . opens upward. Since f has a horizontal shift two units left to x'=-2, then f(-2) provides the vertex of the graph, which is -6. This means x=-2 is the axis of symmetry. This means all the 1 points along the x-axis between -6 and -5 inclusive to the left of the axis of symmetry are part ' of the preimage of [3.10] in R and all the points along the x-axis between 1 and 2 inclusive to 5 the right of the axis of symmetry are also part of the preimage of [3,10] in R . I , Thus, the preimage of [3,10] in IR is {-6,-5] U [1.2]