# 1.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)= 7-e^-x 2.

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Find the inverse for `f(x)=7-e^(-2x)`

The function `f(x)=7-e^(-2x)` is a one-to-one function, so it has an inverse on its domain. The domain is all real numbers. The range is `(-oo,7)` . (The range of `e^(-x^2)` is `(0,oo)` ; it is an exponential decrease function.)

One method to find the inverse is to exchange the x and y, and then solve for y.

`y=7-e^(-2x)` is the original function.

`x=7-e^(-2y)` will give the inverse relation. Solving for y we get:

`e^(-2y)=7-x`

`ln(e^(-2y))=ln(7-x)`

`-2y=ln(7-x)`

`y=-1/2 ln(7-x)`

The domain for this function is `(-oo,7)` as expected -- a function's inverse has for its domain the range of the function, and the range of the inverse is the domain of the original function. Thus the range of the inverse is all real numbers.

** The domain of `-1/2ln(7-x)` is restricted only by the logarithm function, whose argument must be greater than zero; thus x must be less than zero for 7-x>0. The range of the logarithm function is all real numbers. **

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**If `f(x)=7-e^(-2x)` then `f^(-1)(x)=-1/2 ln(7-x)` **

**The domain of `f(x)` is all real numbers -- the domain of the inverse is `(-oo,7)` **

**The range of `f(x)` is `(-oo,7)` while the range of the inverse is all real numbers.**

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The graph of `f(x)` in black -- its inverse in red. Note that f(x) has a horizontal asymptote of y=7 as x grows without bound, and that the inverse has a vertical asymptote at x=7:

Note that the function and its inverse are reflections across the line y=x (dotted)