## Inverse Functions

Inverse functions are only defined if the original function is a one to one function, which means that every point on the x axis only corresponds to one point on the y axis.  In fact we have already seen an inverse function in our exponential and log functions. $f(x)=e^{x}\newline f^{-1}(x)=log_{e}(x)$

The inverse function can be found by switching around x and y (f(x)). $y=e^{x} \newline x=e^{y}$

Take logs of both sides $log_{e}(x)=y=f^{-1}(x)$

Here is the graph of the two functions: The inverse is a reflection of the original function in the black line y=x.

When the function is flipped to the inverse function the new domain becomes the old range and the old domain becomes the new range.

• Dom f-1=ran f
• Ran f-1=dom f

## Example

2014 exam 2 We know that there can only be an inverse function if the function is a one to one function.

For f(x):

• $x \rightarrow infty^{+}, f(x) \rightarrow infty^{+}$
• $x \rightarrow infty^{-}, f(x) \rightarrow infty^{-}$

If a cubic function has a point of inflection then it will be a one to one function, therefore our first step is to find the derivative. $f(x)=2x^{3}-9x^{2}-168\newline f'(x)=6x^{2}-18x-168$

Now we solve for x in f'(x)=0. $f'(x)=6x^{2}-18x-168=0 \newline 6((x-7)(x+3))=0$

Therefore we have two turning points and it is not a one to one function, so we need to restrict the domein.

There are three sections of the graph where it is a one to one function

• D=(∞, -3)
• D=( -3,7)
• D=(7,∞+)

Only one of those are the same as the above which is the D:(∞, -3) which is A in the multiple choice.