For f(x) = 2e^(x-2)+1, what are the transformations from e^x. Also, what are the domain and range of f(x). Find f^-1 (x) and its domain and range.

Question

For f(x) = 2e^(x-2)+1, what are the transformations from e^x. Also, what are the domain and range of f(x). Find f^-1 (x) and its domain and range.

Answer 1

The parent equation for f(x) = 2e^x-2 + 1 is g(x) = e^x.

 

The function has been shifted up 1 (from + 1), shifted right 2 (from - 2), and stretched vertically by a factor of 2 (from the first 2).

 

The domain of the function is all real numbers, since there are no restrictions on x in e^x. The range of y = e^x is all numbers greater than 0, so since the function has been shifted up 1, the new range of the function is all numbers greater than 1.

 

 

 

To find f^-1(x), swap x and y and then solve for y:

 

y = 2e^x-2 + 1

x = 2e^y-2 + 1

(x - 1)/2 = e^y-2

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

ln((x-1)/2) + 2 = y

 

f^-1(x) = ln((x-1)/2) + 2

 

Since the argument of any log must be greater than 0, (x-1)/2 > 0. Thus x - 1 > 0, so x > 1. The domain of the inverse function is all numbers greater than 1. That's the same as the range of the original function.

 

The range of the normal log function y = log(x) is all real numbers. That's the same as the range of the inverse function: all real numbers. Notice that it's also the same as the domain of the original function.

 

So, to find the domain and range of the inverse, you just have to switch the domain and range of the original.