find the composite function fog
f(X) = 1/X-5,g(x)-6/x
question 2
find inverse, domain range and asymatotes of each function
f(x)=3+e^4-x
can some one help me please been stuck on these for 2 days need help solving thank you
assuming a typo, I see
f(x) = 1/(x-5)
g(x) = 6/x
so, f◦g = f(g) = 1/(g-5)
= 1/((6/x)-5)
= 1/((6-5x)/x)
= x/(6-5x)
check for some value, say, x=2:
g(2) = 3
f(3) = -1/2
(f◦g)(2)=2/(6-5*2) = 2/-4 = -1/2
y=3+e^(4-x)
y-3 = e^(4-x)
ln(y-3) = 4-x
x = 4 - ln(y-3)
so, f^-1(x) = 4 - ln(x-3)
domain of f is all reals
range is y>3
asymptote is y=3
Sure, I'd be happy to help you with these questions. Let's start with the first one: finding the composite function fog.
The composite function fog is obtained by plugging g(x) into f(x). In other words, instead of using the variable x in f(x), we will use the expression g(x).
Given:
f(x) = 1/(x - 5)
g(x) = 6/x
To find fog, substitute g(x) into f(x):
fog = f(g(x))
Replace the x in f(x) with g(x):
fog = 1/(g(x) - 5)
Now, replace g(x) with its value:
fog = 1/((6/x) - 5)
To simplify this expression, we first need to find a common denominator for the fraction in the denominator:
fog = 1/((6 - 5x)/x)
Then, invert the denominator and multiply:
fog = x/(6 - 5x)
So, the composite function fog is equal to x/(6 - 5x).
Now, let's move on to the second question: finding the inverse, domain, range, and asymptotes of the function f(x) = 3 + e^(4 - x).
To find the inverse of f(x), we need to switch the roles of x and y. Let's solve for y:
y = 3 + e^(4 - x)
Now, swap x and y:
x = 3 + e^(4 - y)
Next, solve for y:
x - 3 = e^(4 - y)
Taking the natural logarithm of both sides:
ln(x - 3) = 4 - y
Rearranging the equation:
y = 4 - ln(x - 3)
So, the inverse of f(x) is given by y = 4 - ln(x - 3).
For the domain, we need to consider any restrictions on the values x can take. In this case, the natural logarithm function has a restriction that the argument (x - 3) must be greater than zero. So the domain of f(x) is x > 3.
The range of f(x) can include all possible values that the function can output. Since e^(4 - x) is always positive, the minimum value of f(x) is 3. Thus, the range of f(x) is y ≥ 3.
Regarding asymptotes, we need to consider both horizontal and vertical asymptotes.
Horizontal asymptote:
As x approaches positive or negative infinity, the expression e^(4 - x) becomes negligible compared to 3. Therefore, the horizontal asymptote of f(x) is y = 3.
Vertical asymptote:
There are no vertical asymptotes for this function since it does not have any value that causes the denominator (x - 3) to be zero.
I hope this explanation helps you understand how to find the composite function and determine the inverse, domain, range, and asymptotes of a given function. Let me know if you have any further questions!