Let f:R→]1,+∞[ with f(x)=e^x +1 (Plot the curve of f and f^{-1} in the same graph.)

Steps to Solve

1. Identify the function and its inverse

The given function is ( f(x) = e^x + 1 ).

To find the inverse, replace ( f(x) ) with ( y ): $$ y = e^x + 1 $$

Now, solve for ( x ): $$ y - 1 = e^x $$

Take the natural logarithm of both sides: $$ \ln(y - 1) = x $$

Thus, the inverse function is: $$ f^{-1}(y) = \ln(y - 1) $$

2. Define the domains

For the function ( f(x) = e^x + 1 ):

• The domain is all real numbers ( ( -\infty, +\infty ) ).
• The range is ( (1, +\infty) ) since ( e^x ) is always positive and ( e^x + 1 > 1 ).

For the inverse function ( f^{-1}(y) = \ln(y - 1) ):

• The domain is ( (1, +\infty) ) since ( y ) must be greater than 1.
• The range is all real numbers ( ( -\infty, +\infty ) ).

3. Plot the original function and its inverse

To plot ( f(x) = e^x + 1 ):

• Choose values for ( x ) (e.g., -2, -1, 0, 1, 2).
• Calculate corresponding ( f(x) ):
  • ( f(-2) = e^{-2} + 1 )
  • ( f(-1) = e^{-1} + 1 )
  • ( f(0) = e^0 + 1 = 2 )
  • ( f(1) = e + 1 )
  • ( f(2) = e^2 + 1 )

For the inverse function ( f^{-1}(y) ):

• Choose values for ( y ) within its domain (e.g., 2, 3, 4, 5).
• Calculate the corresponding ( f^{-1}(y) ):
  • ( f^{-1}(2) = \ln(2 - 1) = \ln(1) = 0 )
  • ( f^{-1}(3) = \ln(3 - 1) = \ln(2) )
  • ( f^{-1}(4) = \ln(4 - 1) = \ln(3) )
  • ( f^{-1}(5) = \ln(5 - 1) = \ln(4) )

4. Graph both functions

Use a graphing tool or plot on graph paper:

• Plot points obtained from step 3 for both functions.
• Connect the points smoothly.

5. Analyze the graph

Check the symmetry of the graphs about the line ( y = x ) which is a property of functions and their inverses.