Could someone please check my work

2. Find f -([3, 10]): [-6,-5] v [1,2] Explain: f -'([3, 10]) = {x E R: f(x)) E [3, 10]} = [ - 6, - 5] U [1, 2] by notation 2.3.13 for the preimage of [3, 10] ER. Taking the inverse of f(x) means x = (f -1(a) + 2)- 6, which means x+6 = (f 1(x) + 2) .Squaring both sides Iberoblanco Jon idi on godmun w set outun lsat s blond (vx 46) = f-1(x) + 2, which means er sam noirtwi s f (ac) = | vac + 6| - 2. Then substituting x=3 into f-1(X) means some yin uni Sex gobut long -1(3) = |V3+ 6- 2, which means andom fairly . - f -1(3) = 131 - 2 , which means ansem of -1(3) = { - 5, 1} . Now substituting x=10 into f-1(x) means f -1(10) = /v10 + 6 - 2, which means of1 20bivong (S ) aws & co f51(10) = 14| - 2 , which means some .bisway pogo He ansem ault spoontiny: to aixs e Wifeange le aly sri, Bos (f T (10) = { - 6, 2} . noswed eixe-x eill gnole cliing 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 -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 the right of the axis of symmetry are also part of the preimage of [3, 10] in R . Thus, the preimage of [3, 10] in R is [-6,-5] U [1,2]