17..中四MATH..請詳列步驟..

若f(x)=x^2+x及g(x)=2x+1..
求以下各題的值
(a)
f(2)
=4+2
=6
g(2)
=4+1
=5
f(2)+g(2)
=6+5
=11
b)
f(-1)
=1-1
=0
g(-1)
=-2+1
=-1
f(-1)-g(-1)
=0-(-1)
=1
c)
f(1/3)
=(1/3)^2+1/3
=4/9
g(1/3)
=2/3+1
=5/3
f(1/3)．g(1/3)
=(4/9)(5/3)
=20/27
d)
f(-1/4)
=(-1/4)^2-1/4
=1/16-1/4
=-3/16
g(-1/4)
=2(-1/4)+1
=-1/2+1
=1/2
g(-1/4)／f(-1/4)
=(1/2)/(-3/16)
=-8/3

a) f(2)+g(2)
=(2^2+2)+(2*2+1)
=4+2+4+1
=11

b) f(-1)-g(-1)
=[(-1)^2+(-1)]-[2(-1)+1]
=1-1+2-1
=1

c) f(1/3)．g(1/3)
=[(1/3)^2+1/3]*[2(1/3)+1]
=(1/9+1/3)(2/3+1)
=(1/9+3/9)(2/3+3/3)
=4/9*5/3
=20/27

d) g(-1/4)／f(-1/4)
=[2(-1/4)+1]/[(-1/4)^2+(-1/4)]
=(-1/2+1)/(1/16-1/4)
=(-1/2+2/2)/(1/16-4/16)
=(1/2)/(-3/16)
=-1/2*16/3
=-8/3

a)f(2)=2^2+2
g(2)=2(2)+1
f(2)+g(2)=2^2+2+2(2)+1
=6+5
=11

b)f(-1)=(-1)^2+(-1)
g(-1)=2(-1)+1
f(-1)-g(-1)= (-1)^2+(-1)-2(-1)+1
=0-(-1)
=1


c) f(1/3)．g(1/3)
f(1/3)=(1/3)^2+(1/3)
g(1/3)=2((1/3)+1
f(1/3) ‧g(1/3)=[(1/3)^2+(1/3)]‧[(1/3)+1]
=4/9‧4/3
=16/9

d) g(-1/4)／f(-1/4)
f(-1/4)=( -1/4)^2+(-1/4)
g(-1/4)=2((-1/4)+1
g(-1/4)/(-1/4) =[(-1/4)+1] /[(-1/4)^2+(-1/4)]
=(3/4)/(5/16)
=12/5

a) f(2)=4+2=6, g(2)=4+1=5
f(2)+g(2)=6+5=11

b) f(-1)=1-1=0, g(-1)=-2+1=-1
f(-1)-g(-1)=0+1=1

c) f(1/3)=1/9+1/3=4/9, g(1/3)=2/3+1=15/9
f(1/3)+g(1/3)=4/9+15/9=19/9

d)g(-1/4)=1/16-1/4=-3/16 ,g (-1/4)=-2/4+1=2/4=8/16
3/16 x 16/8=1/8