# function is onto as every horizontal line | My Assignment Tutor

1.(i) Prove f is one-to-one.Let f(x1) = f(x2). We need to show that x1 = x2. Suppose the contrary. x1 6= x2 )))Contradiction. Therefore, f is injective.Prove f is not onto.Let f(x) = 5 2 R = ran f. Thenx2 1 6= x2 2 (because of the nonnegative domain)x2 1 + 1 6= x2 + 1f(x1) 6= f(x2) x2 + 1 = 5×2 = 4x = 2 62 dom fTherefore, f is not surjective.Change the range and (optional) find f-1.Notice that for any x 2 dom f = [0; 1); x2 2 [0; 1). So x2 + 1 2 [1; 2). If we set ran f = [1; 2), then f issurjective and f-1 exists. To find it, interchange x and y and solve for y.x = y2 + 1y2 = x – 1y = px – 1Therefore, f-1 : [1; 2) ! [0; 1); f-1(x) = px – 1.(ii) Prove f is not one-to-one.Let f(x1) = f(x2) = 16. Since (-2)4 = 24 = 16 and f-2; 2g ⊆ dom f, we have 2 different domain elementsthat give the same range element. Therefore, f is not injective.Prove f is onto.Let f(x) = y 2 ran f. Thenx4 = y±p4 x4 = p4 y (y ≥ 0)±x = p4 y 2 R = dom fSo for every y 2 ran f, there exists x 2 dom f such that f(x) = y. Therefore, f is surjective.Change the range and (optional) find f-1.We can see from the disproof of injectivity that restricting dom f to R+ (or to R-) makes f injective andguarantees the existence of f-1. Find it by interchanging x and y.x = y4y = p4 xTherefore, f-1 : R+ ! R+; f-1(x) = p4 x.2. (i) The function is one-to-one as the graph of f satisfies the horizontal line test (also f(x1) = f(x2) ) x1 = x2).The function is onto as every horizontal line meets the graph (also given y 2 R there is x = p3 y such thatf(x) = y). Hence the function has an inverse f-1 : R ! R given by f-1(x) = p3 x.1(ii) Since we know f-1 exists, we find it by standard means:x =y1 – yx(1 – y) = yx – xy = y-y – xy = -x-y(1 + x) = -xy =x1 + xTherefore, f-1 : R++ ! (0; 1); f-1(x) = x+1 x .(iii) The function cos : R ! R is not one-to-one as cos 0 = cos 2π and not onto as the value 2 is not taken. If werestrict the domain to [0; π] (or similar), then cos : [0; π] ! [-1; 1] is one-to-one (passes horizontal line test)and onto (every horizontal line cuts the graph), Hence, the inverse arccos : [-1; 1] ! [0; π] can be defined.(iv) The function tan : R ! R is not one-to-one as tan 0 = tan 2π but it is onto (every horizontal line cuts thegraph). If we restrict the domain to -π2 ; π2 (or similar), then tan : -π2 ; π2 ! R is one-to-one and onto andhence the inverse arctan : R ! -π2 ; π2 can be defined.(v) The function is not onto as the value -1 is not taken. If we restrict the range to (0; 1) then the functionis one-to-one (passes horizontal line test) and onto (every horizonal line cuts the graph). Hence the inverseln : (0; 1) ! R can be defined.3. There are several answers; for instance, VH = fv1g, EH = fe1; e2g.4. (i) .(ii) .(iii) .(iv) A simple graph with 5 vertices including one of degree 5 is not possible.(v) Not possible; 5 edges means the graph has degree 10, but the sum of vertex degrees is 8.(vi) .5. .2

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