posted by elley on .
show that the solution has exactly one solution in the interval
x+ln(x+1)=0, 0<and equal to x <and equal to 3
x+ln(x+1)=0 , 0 ≤ x ≤ 3
write your equation as ln(x+1) = -x
now let the left side by
y = ln(x+1) and the right side
y = -x
and make a quick sketch of each on the same x-y plane.
the first one is the log function crossing the x axis at the origin, slowing rising in the first quadrant, and dropping down into the third quadrant.
the other graph is a straigh line with slope of -1 and passing through the origin.
Clearly they can only cross at x=0, which becomes your only answer.
check : if x=0
LS = 0 + ln(0+1)
= 0 + 0 = 0 = RS