Consider the logarithmic equation f(x) = ln (x + 2) – 1. Round all values to the tenths place.
1 answer:
Hello,
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
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