find the domain

[logistic_guy]

here is the question

Find the domain of the function $\displaystyle f(x) = \frac{x + 1}{\ln(1 - 2x)}$.


my attemb
this function is my enemy. this isn't the first time to work with it
i'll always get the domain worng and now i think it's the best moment to show
i'm very good in domain after studying $\displaystyle 6$ treatments of composition functions
i'll try to solve it
i've fraction so i don't want the denomator to be zero
$\displaystyle \ln(1 - 2x) = 0$
$\displaystyle 1 - 2x = e^0 = 1$
$\displaystyle -1 + 2x = -1$
$\displaystyle 2x = -1 + 1 = 0$
$\displaystyle x = 0$
so zero is invalid value in the domain
my think now i don't want to have negative inside the logarthm function
so i want $\displaystyle 1 - 2x > 0$
$\displaystyle -1 + 2x < 0$
$\displaystyle 2x < 1$
$\displaystyle x < \frac{1}{2}$
Domain: $\displaystyle (0,\frac{1}{2})$

the graph of $\displaystyle f(x)$ don't agree with my calculations

 

[Dr.Peterson]

here is the question

Find the domain of the function $\displaystyle f(x) = \frac{x + 1}{\ln(1 - 2x)}$.


my attemb
this function is my enemy. this isn't the first time to work with it
i'll always get the domain worng and now i think it's the best moment to show
i'm very good in domain after studying $\displaystyle 6$ treatments of composition functions
i'll try to solve it
i've fraction so i don't want the denomator to be zero
$\displaystyle \ln(1 - 2x) = 0$
$\displaystyle 1 - 2x = e^0 = 1$
$\displaystyle -1 + 2x = -1$
$\displaystyle 2x = -1 + 1 = 0$
$\displaystyle x = 0$
so zero is invalid value in the domain
my think now i don't want to have negative inside the logarthm function
so i want $\displaystyle 1 - 2x > 0$
$\displaystyle -1 + 2x < 0$
$\displaystyle 2x < 1$
$\displaystyle x < \frac{1}{2}$
Domain: $\displaystyle (0,\frac{1}{2})$
View attachment 38991
the graph of $\displaystyle f(x)$ don't agree with my calculations

Everything you said was correct until the last statement. What is the set of numbers less than 1/2 and not equal to 0? It is not (0, 1/2).

When you find that your answer appears to be wrong, you don't just cry and ask for help. You check your work to see where the mistake was (and also check the check, in case it was wrong).

 

[Steven G]

What would your answer be if you did not have to exclude x=0? Now exclude x from that answer.

 

[Steven G]

What would your answer be if you did not have to exclude x=0? Now exclude x from that answer.

...now exclude x=0

 

[logistic_guy]

Everything you said was correct until the last statement. What is the set of numbers less than 1/2 and not equal to 0? It is not (0, 1/2).

thank

i miss the idea to check negative numbers. if i do i'll catch the domain $\displaystyle (-\infty, 0) \cup (0, \frac{1}{2})$

When you find that your answer appears to be wrong, you don't just cry and ask for help. You check your work to see where the mistake was (and also check the check, in case it was wrong).

i cry because i thought i'm mastering domain

What would your answer be if you did not have to exclude x=0? Now exclude x from that answer.

i do. i see the idea. appreciate it

...now exclude x=0