Could someone please check my work

Lu insnit ta alarm) 67 Reninh c 3. Find f'1((10, 2)): mi?) 7 Explain: W '2 \"MANU'WD f\"1((10, -2)) ={$R=f(w)) El10, 2)}= [*6, *2) by notation 2.3.13 for the preimage of [ 10, 2] E R . 2 Taking the inverse of f(:v) means cc = (f\"1(m) + 2) 6 , which means a: + 6 = (f_1(:z:) + 2)2 .Squaring both sides Na: + 6] = f\"1(:c) + 2 , which means f\"1(a:) = Na: + 6[ - 2 . Then substituting x=-10 into f'1(x) means fdl '* 10) = ]\\/10 +6] 2 ,which means f_1( - 10) = lv4l 2, which means f 1( - 10) = [22'] 2 .- This value is not in the set of real K i . . . . . \\a. . numbers, so it is not conSIdered part of the preimage which must be a subset of the real ' numbers. The next largest negative value in {-10,-2) that will yield a real number is -6, which i is the vertex of f, so substituting x=-6 into f'1(x) means ' ' f'1(6) = [W] -2,whichmeans ' f \"1( - 6) : 2 . Then substituting x=-2 into f\"(x) means f\"1(-2) = IWILwhichmeans f'1( 2) = l2] 2 , which means f'1(-2) = {0, 4}. i The graph of f is a parabolic function. Since the graph of f has a positive coefficient of x, f i 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 xn-Z is the axis of symmetry. This means all the j points along the x-axis between -4 (but not inclusive of -4) and the axis of symmetry (inclusive of the axis of symmetry) are part of the preimage of [40.4] in R and all the points along the x-axis between 0 (but not inclusive of O) and the axis of symmetry (inclusive of the aXIS of symmetry) are also part of the preimage of [3,10] in IR . Thus, the preimage of (-10,-2) in IR is (0,-4)