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43.The graph ofy=x2on the window-10≤x≤10,-10≤y≤10 appears identical (except forlabels) to the graph ofy= 2(x-1)2+ 3 if thelatter is drawn on a graphing window centeredat the point (1,3) with 1-5√2≤x≤1+5√2,-7≤y≤13.44.The graph ofy=x4is below the graph ofy=x2when-1≤x≤1, and above it whenx >1. Both graphs have roughly the same up-ward parabola shape, buty=x4is flatter atthe bottom.45.py2is the distance from (x, y) to thex-axisqx2+ (y-2)2is the distance from (x, y) tothe point (0,2). If we require that these be thesame, and we square both quantities, we havey2=x2+ (y-2)2y2=x2+y2-4y+ 44y=x2+ 4y=14x2+ 1In this relation, we see thatyis a quadraticfunction ofx. The graph is commonly knownas a parabola.46.The distance between (x, y) and the x-axis ispy2.The distance between (x, y)and (1,4)isq(x-1)2+ (y-4)2.Setting these equaland squaring both sides yieldsy2= (x-1)2+(y-4)2which simplifies toy=18(x-1)2+ 16(a parabola).0.3Inverse Fuctions1.f(x) =x5andg(x) =x1/5f(g(x)) =fx1/5=x1/55=xg(f(x)) =g(x5)=(x5)1/5=x(5/5 )=x2.f(x) = 4x3andg(x) =14x1/3f(g(x)) = 414x1/3!3= 414x=xg(f(x)) =144x31/3=x3.f(x) = 2x3+ 1 andg(x) =3rx-12f(g(x)) = 23rx-12!3+ 1= 2x-12+ 1 =xg(f(x)) =3rf(x)-12=3r2x3+ 1-12=3√x3=x4.f(x) =1x+ 2andg(x) =1-2xxf(g(x)) =11-2xx+ 2=11-2xx+2xx=xg(f(x)) =1-21x+21x+2=1-21x+ 2(x+ 2)= (x+ 2)-2 =x5.The function is one-to-one sincef(x) =x3isone-to-one. To find the inverse function, writey=x3-2y+ 2 =x33py+ 2 =xSof-1(x) =3√x+ 2-110-5-24354x0-12-41-3y-32-2-43-556.The function is one-to-one with inversef-1(x) =3√x-4
0.3.INVERSE FUCTIONS15-48-842y0-64-1060-610-210-48-2x-826-107.The graph ofy=x5is one-to-one and henceso isf(x) =x5-1. To find a formula for theinverse, writey=x5-1y+ 1 =x55py+ 1 =xSof-1(x) =5√x+ 1y04-5-3-343552-1-21-4-5-12-203-41x8.The function is one-to-one with inversef-1(x) =5√x-4.50-52-3-3254x-4-10-21-5-1-23y1-4349.The function is not one-to-one since it is aneven function (f(-x) =f(x)).In particular,f(2) = 18 =f(-2).10.Not one-to-one. Fails horizontal line test.11.Here, the natural domain requires that theradicand (the object inside the radical) benonnegative. Hencex≥ -1 is required, whileall function values are non negative. Thereforethe inverse, if defined at all, will be definedonly for nonnegative numbers. Sometimes onecan determine the existence of an inverse inthe process of trying to find its formula. Thisis an example: Writey=px3+ 1y2=x3+ 1y2-1 =x33py2-1 =xThe left side is a formula forf-1(y), good fory≥0.Therefore,f-1(x) =3px2-1 when-everx≥0.021-2-2332-1x-1y1012.Not one-to-one. Fails horizontal line test.13.(a) Sincef(0) =-1, we knowf-1(-1) = 0(b) Sincef(1) = 4, we knowf-1(4) = 114.(a) Sincef(0) = 1, we knowf-1(1) = 0.