How do you find the domain of #f(x)=3/(1+e^(2x))#?

Question

Kennedy Adams · Accepted Answer

Domain: #x in RR#

(defined for all values of #x# in the real set...)

To answer this problem, the first thing we can consider is if there are any values, for what the function, #f(x) # is undifined at, this would be where the denominator #=0#

#=> 1 + e^(2x) = 0 #

#=> e^(2x) = -1 #

#=> 2x = ln(-1) #

#=> x = 1/2 ln(-1) #

We know #1/2 ln(-1) notin RR #

So hence the denominator is defined #AAx in RR #

( defined for all #x# values in the real set)

We also know #e^(2x)# is also defined #AAx in RR#

So hence #f(x) # is defined for #AAx in RR #

Domain: #x in RR #

(for all real values of #x# )

Ethan Adkins · Answer

undefined To find the domain of the function f(x) = 3/(1 + e^(2x)), note that the function will be undefined where the denominator is equal to zero, since division by zero is undefined. Therefore, the domain consists of all real numbers except those values that make the denominator equal to zero. In this case, the denominator is 1 + e^(2x). Setting this expression equal to zero and solving for x gives:

1 + e^(2x) ≠ 0

e^(2x) ≠ -1

Since the exponential function e^(2x) is always positive for real values of x, it can never equal -1. Therefore, there are no values of x that make the denominator zero. Hence, the domain of the function f(x) = 3/(1 + e^(2x)) is all real numbers, or (-∞, ∞).