I think you mean f(x) = 3 + e^(4−x)
To get the inverse, replace the f(x) by x, and the x by f'(x), then solve for f'(x):
f(x) = 3 + e^(4−x)
x = 3 + e^(4 - f'(x))
x - 3 = e^(4 - f'(x))
ln(x - 3) = 4 - f'(x)
f'(x) = 4 - ln(x - 3)
This is the inverse of the original function.
i. Domain
Domain is the set of all possible values of x.
f(x) = 3 + e^(4−x)
For this function, the domain is all real numbers.
Now, try to determine the domain of the inverse function, f'(x) = 4 - ln(x - 3)
ii. Range
Range is the set of all possible values of f(x).
f(x) = 3 + e^(4−x)
For this function, the range is all real numbers greater than 3. Note that the smallest value that e^(4-x) approaches, is zero, which happens if x is very large.
Now, try to determine the range of the inverse function, f'(x) = 4 - ln(x - 3)
iii. Asymptote
To get the horizontal asymptote, we get the limit of the function as x -> infinity.
To get the vertical asymptote, we get the limit of the function as f(x) -> infinity.
f(x) = 3 + e^(4−x)
Horizontal Asymptote: f(x) = 3
Vertical Asymptote: none
Now, try to determine the asymptotes of the inverse function, f'(x) = 4 - ln(x - 3)
Hope this helps :3
Find the inverse of the following function. Find the domain, range, and asymptotes of each function. Graph both functions on the same coordinate plane.
f(x)=3+e^4−x
2 answers
I will assume you meant
f(x) = 3 + e^(4-x)
or else you would just have a straight line
since e^? can have any real number as an exponent,
the domain of the function is any real number
for the range ...
as x ---> positive large, e^(4-x) becomes very small
e.g. e^-100 = 3.7x10^-44
the function approaches 3 + 0 which is 3
as x ---> negative large, e^(4-x) approaches infinitiy.
so for the original f(x)
domain: any real number
range: any real number, y>3
inverse:
step 1: for y = 3 + e^(4-x), interchange the x and y variables to get
x = 3 + e^(4-y)
x-3 = e^(4-y)
take ln of both sides
ln(x-3) = ln e^(4-y)
ln(x-3) = 4-y
y = 4 - ln(x-3)
of course the domain of the original becomes the range of the inverse
and the range of the original becomes the domain of the inverse
f(x) = 3 + e^(4-x)
or else you would just have a straight line
since e^? can have any real number as an exponent,
the domain of the function is any real number
for the range ...
as x ---> positive large, e^(4-x) becomes very small
e.g. e^-100 = 3.7x10^-44
the function approaches 3 + 0 which is 3
as x ---> negative large, e^(4-x) approaches infinitiy.
so for the original f(x)
domain: any real number
range: any real number, y>3
inverse:
step 1: for y = 3 + e^(4-x), interchange the x and y variables to get
x = 3 + e^(4-y)
x-3 = e^(4-y)
take ln of both sides
ln(x-3) = ln e^(4-y)
ln(x-3) = 4-y
y = 4 - ln(x-3)
of course the domain of the original becomes the range of the inverse
and the range of the original becomes the domain of the inverse
2 answers