5. Let f: R+R be the function defined by f(x) = 2x4 – 4 for any z € IR. (a) Fill in the blanks in the block below, all l

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5 Let F R R Be The Function Defined By F X 2x4 4 For Any Z Ir A Fill In The Blanks In The Block Below All L 1 (29.28 KiB) Viewed 96 times

5. Let f: R+R be the function defined by f(x) = 2x4 – 4 for any z € IR. (a) Fill in the blanks in the block below, all labelled by capital-letter Roman numerals, with appropriate words so that it gives a proof for the equality f([1,2]) = (-2, 28). (The 'underline' for each blank bears no definite relation with the length of the answer for that blank.)
Write S = [1,2]. . [We want to verify the statement (t): 'for any y, if ye f(S) then y € (-2, 28).] (I) Pick any y. Then by the definition of f(S), (II) For the same r, since z ES, we have 1<x< 2. Since x > 1, we have (III) (IV) we have y = f(x) = 2x* - 4<2-24 - 4 = 28. Therefore -2 <y < 28. Hence y € (-2,28]. We want to verify the statement (1): 'for any y, if y € (-2, 28) then y e f(S).') Pick any y. Suppose y € (-2,28]. Then -2y < 28. [We want to verify that for this y, there exists some x € S such that y=f(x).] (V) We verify that x ES: (VI) * Since y >-2, we have y +4 > 1. Then 2 (VII) Then Therefore 1 <r < 2. Hence x € [1, 2] = S. (VIII) For the same r, we have (IX) Then, for the same x, y, we have reS and y = f(x). Hence by the definition of f(S), (X) It follows that f(S) = (-2,28). (b) Fill in the blanks in the block below, all labelled by capital-letter Roman numerals, with appropriate words so that it gives a proof for the equality f-(-6,4]) = (-V2, 2). (The 'underline' for each blank bears no definite relation with the length of the answer for that blank.)
Write U = (-6,4]. [We want to verify the statement (t): 'for any z, if x e f-'(U) then x € [-V2, V2].] Pick any . (1) . E Then by the definition of f-1(U), (II) For the same y, since (III) we have -6 <y < 4. Since y> -6, we have 2x4 -4 = f(x) = y >-6. Then x4> -1. (This provides no information other than re-iterating 'x e R') Since y < 4, we have (IV) Then x4 < 4. Since x E IR, we have -V2<x< V2. Then x € [-V2, V2]. . [We want to verify the statement (1): 'for any z, if x € [-V2, V2] then x E F-1(U).!] Pick any x. (V) Then -V2<x< V2. [We want to verify that for this x, there exists some y E U such that y = f(x).] (VI) We verify that y EU: (VII) * Since - V2 <<<V2, we have x4 < 4. Then Since x e R, we have x4>02-1. Then Therefore -6 <y < 4. Hence y € (-6, 4] =U. (VIII) Then, for the same x, y, we have y = f(x) (IX) YEU. Hence by the definition of f-1(U), (X) It follows that f-1(U) = (-V2, V2].