# MATH 103 Solutions - Section 1.1 C01S01.001: If f x = 1...

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Section 1.1C01S01.001:Iff(x) =1x, then:(a)f(-a) =1-a=-1a;(b)f(a-1) =1a-1=a;(c)f(a) =1a=1a1/2=a-1/2;(d)f(a2) =1a2=a-2.C01S01.002:Iff(x) =x2+ 5, then:(a)f(-a) = (-a)2+ 5 =a2+ 5;(b)f(a-1) = (a-1)2+ 5 =a-2+ 5 =1a2+ 5 =1 + 5a2a2;(c)f(a) = (a)2+ 5 =a+ 5;(d)f(a2) = (a2)2+ 5 =a4+ 5.C01S01.003:Iff(x) =1x2+ 5, then:(a)f(-a) =1(-a)2+ 5=1a2+ 5;(b)f(a-1) =1(a-1)2+ 5=1a-2+ 5=1·a2a-2·a2+ 5·a2=a21 + 5a2;(c)f(a) =1(a)2+ 5=1a+ 5;(d)f(a2) =1(a2)2+ 5=1a4+ 5.C01S01.004:Iff(x) =1 +x2+x4, then:(a)f(-a) =1 + (-a)2+ (-a)4=1 +a2+a4;(b)f(a-1) =1 + (a-1)2+ (a-1)4=1 +a-2+a-4=(a4)·(1 +a-2+a-4)a4=a4+a2+ 1a4=a4+a2+ 1a4=a4+a2+ 1a2;(c)f(a) =1 + (a)2+ (a)4=1 +a+a2;(d)f(a2) =1 + (a2)2+ (a4)2=1 +a4+a8.C01S01.005:Ifg(x) = 3x+ 4 andg(a) = 5, then 3a+ 4 = 5, so 3a= 1; thereforea=13.C01S01.006:Ifg(x) =12x-1andg(a) = 5, then:1
12a-1= 5;1 = 5·(2a-1);1 = 10a-5;10a= 6;a=35.C01S01.007:Ifg(x) =x2+ 16 andg(a) = 5, then:a2+ 16 = 5;a2+ 16 = 25;a2= 9;a= 3 ora=-3.C01S01.008:Ifg(x) =x3-3 andg(a) = 5, thena3-3 = 5, soa3= 8. Hencea= 2.C01S01.009:Ifg(x) =3x+ 25 = (x+ 25)1/3andg(a) = 5, then(a+ 25)1/3= 5;a+ 25 = 53= 125;a= 100.C01S01.010:Ifg(x) = 2x2-x+ 4 andg(a) = 5, then:2a2-a+ 4 = 5;2a2-a-1 = 0;(2a+ 1)(a-1) = 0;2a+ 1 = 0 ora-1 = 0;a=-12ora= 1.C01S01.011:Iff(x) = 3x-2, thenf(a+h)-f(a) = [3(a+h)-2]-[3a-2]= 3a+ 3h-2-3a+ 2 = 3h.C01S01.012:Iff(x) = 1-2x, then2
f(a+h)-f(a) = [1-2(a+h)]-[1-2a] = 1-2a-2h-1 + 2a=-2h.C01S01.013:Iff(x) =x2, thenf(a+h)-f(a) = (a+h)2-a2=a2+ 2ah+h2-a2= 2ah+h2=h·(2a+h).C01S01.014:Iff(x) =x2+ 2x, thenf(a+h)-f(a) = [(a+h)2+ 2(a+h)]-[a2+ 2a]=a2+ 2ah+h2+ 2a+ 2h-a2-2a= 2ah+h2+ 2h=h·(2a+h+ 2).C01S01.015:Iff(x) =1x, thenf(a+h)-f(a) =1a+h-1a=aa(a+h)-a+ha(a+h)=a-(a+h)a(a+h)=-ha(a+h).C01S01.016:Iff(x) =2x+ 1, thenf(a+h)-f(a) =2a+h+ 1-2a+ 1=2(a+ 1)(a+h+ 1)(a+ 1)-2(a+h+ 1)(a+h+ 1)(a+ 1)=2a+ 2(a+h+ 1)(a+ 1)-2a+ 2h+ 2(a+h+ 1)(a+ 1)=(2a+ 2)-(2a+ 2h+ 2)(a+h+ 1)(a+ 1)=2a+ 2-2a-2h-2(a+h+ 1)(a+ 1)=-2h(a+h+ 1)(a+ 1).C01S01.017:Ifx >0 thenf(x) =x|x|=xx= 1.Ifx <0 thenf(x) =x|x|=x-x=-1.We are givenf(0) = 0, so the range offis{-1,0,1}. That is, the range offis the set consisting of thethree real numbers-1, 0, and 1.C01S01.018:Givenf(x) = [[3x]], we see that3
f(x) = 0if0x <13,f(x) = 1if13x <23,f(2) = 2if23x <1;moreover,f(x) =-3if-1x <-23,f(x) =-2if-23x <-13,f(x) =-1if-13x <0.In general, ifmis any integer, thenf(x) = 3mifmx < m+13,f(x) = 3m+ 1ifm+13x < m+23,f(x) = 3m+ 2ifm+23x < m+ 1.Because every integer is equal to 3mor to 3m+ 1 or to 3m+ 2 for some integerm, we see that the rangeoffincludes the setZof all integers. Becausefcan assume no values other than integers, we can concludethat the range offis exactlyZ.C01S01.019:Givenf(x) = (-1)[[x]], we first note that the values of the exponent [[x]] consist of all theintegers and no other numbers. So all that matters about the exponent is whether it is an even integer oran odd integer, for if even thenf(x) = 1 and if odd thenf(x) =-1. No other values off(x) are possible,so the range offis the set consisting of the two numbers-1 and 1.C01S01.020:If 0< x1, thenf(x) = 34. If 1< x2 thenf(x) = 34 + 21 = 55. If 2< x3 thenf(x) = 34 + 2·21 = 76. We continue in this way and conclude with the observation that if 11< x <12 thenf(x) = 34 + 11·21 = 265. So the range offis the set{34,55,76,97,118,139,160,181,202,223,244,265}.C01S01.021:Givenf(x) = 10-x2, note that for every real numberx,x2is defined, and for every suchreal numberx2, 10-x2is also defined. Therefore the domain offis the setRof all real numbers.

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