Could someone please check my work

3.Findf 'uuo. any Ln. n | Explain: f '(( 10.~2iizireR=I(w))e(~10.~2)}=i6,2) by notation 2.3.13 for the preimage of | 10, ~ 2] E [R . 2 Taking the inverse of at) means a: = (f \"(33) + 2) 6 , which means a: + 6 = (f T 1(3:) + 2)2 .Squaring both sides Na: -+- 6| = f'1(:c) + 2 , which means f " 1(:c) 2 Wm + 6| - 2 . Then substituting x=-10 into f'1(x) means f-1( - 10) = |\\/10 + 6| 2 , which means f1(~ 10) = |\\/Z| - 2 , which means f_1( - 10) = |2i| 2 . This value is not in the set of real K V 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 | is the vertex of f, so substituting x=-6 into f'1(x) means f1(' 6) = Iv6 + 3| ~ 2 , which means f '1( 6) = - 2 . Then substituting x=-2 into f'1(x) means f_1( "' 2) = |v-2 +6| - 2,which means f1(- 2) = I2] - 2,which means f'1(-2) ={0i 4}- 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 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 [-10.-2] in R and all the points along the x-axis between 0 (but not inclusive of 0) and the axis of symmetry (inclusive of the axis of symmetry) are also part of the preimage of [3,10] in R . ' Thus, the preimage of (-10,-2) in R is (0,-4)