MT1710 - 2018/19: Question Sheet 1: Hints to Solutions 1. (a) y = 2-x2 is the parabola with the y-axis as axis and vertex the point (0, 2) and passing through the point (v2,0). (a) y 2 w=2-x2 y=1 1 (b) y 1 4 y =/(2 - x)2 1 22 x 1 2 3 x (b) y = x is the parabola with the x-axis as axis, vertex the origin and passing through the point (1, 1); y = 2 - x line having gradient -1 and passing through the point (1, 1) whilst y = (2 - x)2 is the parabola with vertex the point (2, 0), the line x = 2 as axis and passing through the point (3, 1). (c) By definition, 2x; x?0, x < 0 , with |2-x| = 2 -x; x ?2, x-2; x> 2. -2x; : Hence, we are interested in the three intervals -2 << 0, 0<x?2 and 2<x<3 for which we have y = -2x - (2 -x) = - (x +2); - 2 < x < 0 , 2x-(2-x)=3x-2; 0?x ?2, 2x-(x-2)=x+ 2; 2<x<3. y (c) 5 4 3 2 1 -2 -1 ? ? ? ? ? ? ? ? ? ? ? ? ? 1 2 3 x ? ? ? -1 ? ? ? ? ? 1
2. (a) The function y = f(x) is given by y = { ( x + 2); 2; x ?0. (-(x+2); x< - 2 -2?x<0 The first two may be combined to give y = 1 2; [x+2|; x < 0 x?0. Either strict inequality or weak inequality is acceptable in this example. (b) The function y = f(x) is given by x + 4; x <- 2 -2?x<0 2; v4-x2; 0?x<2 y ={ - x- 2; x > 2. 3. Either working from the diagram, or letting f1(-x) = f1(x) since f1 is even gives: f?(-x) = {1; [ 2- x2 = 2 - (-x)2; 0? x ?1 1?x?2' so that 1; -2<x?-1 fi (x) = {2 - x2; - 1 < x < 1 1 1 << 2 . y 2 y y=2-x2 x=2-x2 1 1 -2 -1 1 2 x -2 -1 1 2 x -1 y = x2 - 2 -2 f1(x) Similarly, letting g1(-x) = - g1(x) since g1 is odd gives: g?(x) J-(2-x2)=(-x)2 - 2; 0< x ?1 g?(-x) =1 -1; 1?x?2' so that ( -1; j {2-x2; 0?x?1 1; 1<x?2 -2?x?-1 - 2; - 1 << < 0 8 . Note that the inequality ? has to be modified to the strict inequality < when discussing the interval [-1, 0), to avoid the f2 taking two distinct values at the origin, thereby violating 2
the definition of a function. Similarly, letting f2(-x) = f2(x) since f2 is even gives: Vã = V-(-xc); 0?x?1 1<x?2, 2?x ?3 f2(-x) =< 2-x=2+(-x); ((2-x)2=(2+(-x))2; so that f2(x) = (2+x)2; -3 5x 5-2 2+ x; -2<x?-1 V-x; -1?x << 0 Vx; 0 << x<1 2 - x; 1<x<2