NAME:Mathematics 144, Fall 2006, Examination 2Answer Key11. [30 points] (a) Is the set of all points (x, y) ∈ R × R such that y2= exthe graphof a function? Give reasons for your answer.(b) Let f : R → R be the function f(x) = 4x + 6. Let A and B be the closed intervals[1, 2] and [−2, 10] respectively. Find the image f[A] and the inverse image f−1[B].(c) Let f : A → B and g : B → C be onto (= surjective) functions. Prove that thecomposite gof is also onto.SOLUTION(a) NO. If y2= exthen also (−y)2= ex, so in particular the set includes the points(0, 1) and (0, −1).(b) The first of these is the set of all points describable as 4x + 6 for 1 ≤ x ≤ 2, whichis merely the interval [10, 14], and the second is the set of all x such that −2 ≤ 4x +6 ≤ 10,which is the interval [−2, 1].(c) Let c ∈ C. Then there is some b ∈ B such that g(b) = c because g is onto.Similarly, since f is onto there is some a ∈ A such that b = f (a). It then follows thatgof(a) = g( f(a) ) = g(b) = c, showing that gof is onto.22. [25 points] (a) If f : [0, ∞) → [0, 1) is the functionf(x) =x2x2+ 1then f is strictly increasing and defines a 1–1 correspondence from its domain to itscodomain, and hence it has an inverse function g. Find a formula for this inverse function.(b) Let f : R → R be the function f(x) = 2x − 1 and let g(x) = x + 1. Evaluategof(2) − fog(2).SOLUTION(a) We need to solve the equationy =x2x2+ 1for x in terms of y. The right hand side can be rewritten as1 −1x2+ 1and therefore we obtain the equation1 − y =1x2+ 1so thatx2+ 1 =11 − yand hencex2=y1 − y.Thus the inverse function is given byg(y) =ry1 − y.(b) We have gof(x) = 2x and fog(x) = 2x + 1. Therefore gof(x) − fog(x) = 1 forall x and in particular for x = 2.33. [20 points] Prove by induction that1 + 3 + · · · + (2n − 1) = n2for all n ≥ 1.SOLUTIONLet Pnbe the staatement displayed above. Then P1is true because both sides areequal to 1. — Now assume that Pkis true; we need to show that Pk+1is also true. Butnow1 + 3 + · · · + (2k + 1) =1 + 3 + · · · + (2k − 1)+ (2k + 1)and by the induction hypothesis (that Pkis true) it follows that the first term on the righthand side is equal to k2. Therefore the right hand side is equal to k2+ 2k + 1 = (k + 1)2,and consequently Pk+1is also true if Pkis true. Therefore, by the weak principle of finieinduction we know that Pnis true for every n ≥ 1.44. [25 points] (a) What is the least upper bound of the closed interval [0, 1] = {x ∈R | 0 ≤ x ≤ 1} in the real numbers? Give reasons for your answer.(b) A well-ordered set is a partially ordered set which satisfies an additional condition.State this condition.SOLUTION(a) The answer is 1. Every element in the set is less than or equal to 1, so 1 is anupper bound. On the other hand, since 1 belongs to the set we also know that any upperbound u must be greater than or equal to 1.(b) Every nonempty subset has a least