Find the inverse of the function below. Graph the function below and the inverse function. Determine the domain, range, and asymptotes of the function below and the inverse function. Please show all of your work. and graph.

f(x)=e^-x/2+4

How about some parens for clarity?

e^(-x/2+4)
e^(-x/2) + 4
(e^(-x))/2 + 4

To find the inverse of a function, we need to interchange the x and y variables and solve for y.

Given function: f(x) = e^(-x/2) + 4

1. Interchange x and y: x = e^(-y/2) + 4
2. Solve for y: Subtract 4 from both sides and take the natural logarithm on both sides to isolate y.
x - 4 = e^(-y/2)
ln(x - 4) = ln(e^(-y/2))
ln(x - 4) = -y/2
Multiply by -2:
-2ln(x - 4) = y

So, the inverse function is: f^-1(x) = -2ln(x - 4).

Now, let's proceed with graphing the original function f(x) and its inverse f^-1(x).

Graph of f(x) = e^(-x/2) + 4:
To graph the original function, we can start by plotting a few points and then determine its general shape. Let's choose some x-values and find the corresponding y-values by substituting them into the equation.

When x = 0: f(0) = e^(0/2) + 4 = 1 + 4 = 5. So, we have the point (0, 5).
When x = 2: f(2) = e^(-2/2) + 4 = e^(-1) + 4 ≈ 0.369 + 4 ≈ 4.369. So, we have the point (2, 4.369).
When x = -2: f(-2) = e^(2/2) + 4 = e^(-1) + 4 ≈ 0.369 + 4 ≈ 4.369. So, we have the point (-2, 4.369).

Based on these points, we can sketch a smooth curve that increases as x approaches negative infinity and approaches the asymptote, which we will determine later.

Graph of f^-1(x) = -2ln(x - 4):
For the inverse function, we can follow a similar process, plotting points and sketching the graph. Let's choose some values for x and find the corresponding y-values.

When x = 0: f^-1(0) = -2ln(0 - 4) = -2ln(-4). Since the natural logarithm is undefined for negative values, the function is not defined for x < 4. The domain of the inverse function is x ≥ 4.
When x = 2: f^-1(2) = -2ln(2 - 4) = -2ln(-2). Again, the logarithm is undefined for negative values, so there is no corresponding point.
When x = 6: f^-1(6) = -2ln(6 - 4) = -2ln(2) ≈ -2(0.693) ≈ -1.386. So, we have the point (6, -1.386).

Based on these points, we can sketch a smooth curve that approaches the asymptote and increases as x approaches positive infinity.

Domain and Range:
For the original function f(x) = e^(-x/2) + 4, as exponential functions are defined for all values of x: The domain is all real numbers, (-∞, ∞).
The range can be determined by observing that the exponential function e^(-x/2) is always positive or zero (since x/2 ≥ 0), and adding 4 to it will result in values at or above 4. Therefore, the range is [4, ∞).

For the inverse function f^-1(x) = -2ln(x - 4):
The domain is restricted by the natural logarithm, which is only defined for positive values. So, the domain is x ≥ 4.
The range can be determined by noting that ln(x - 4) is defined for all positive values. After multiplying by -2, we get negative values, so the range is (-∞, 0).

Asymptotes:
For f(x) = e^(-x/2) + 4, there are no vertical asymptotes since the exponential function is defined for all real numbers. However, there is a horizontal asymptote when x approaches negative infinity, as e^(-x/2) approaches 0. Thus, the horizontal asymptote is y = 4.

For f^-1(x) = -2ln(x - 4), there is a vertical asymptote at x = 4 since the logarithm is undefined for x = 4. There is no horizontal asymptote for the inverse function.

To summarize, the solution includes:
1. The inverse function: f^-1(x) = -2ln(x - 4).
2. Graphs of the original function f(x) = e^(-x/2) + 4 and its inverse f^-1(x).
3. Domain and range:
- Domain of f(x): (-∞, ∞)
- Range of f(x): [4, ∞)
- Domain of f^-1(x): x ≥ 4
- Range of f^-1(x): (-∞, 0)
4. Asymptotes:
- For f(x): Horizontal asymptote at y = 4
- For f^-1(x): Vertical asymptote at x = 4