# Solving inverse function

• Sep 19th 2010, 03:36 PM
iamanoobatmath
Solving inverse function
Hello I am having difficulty solving this question. Any help would be greatly appreciated.
\$\displaystyle g(x)=3+x+e^x\$
The e is the mathematical constant not a variable

find \$\displaystyle g^{-1}(4)\$

So I must find a value of x where g(x)=4

this is my attempt:
\$\displaystyle 4=3+x+e^x\$
\$\displaystyle 1=x+e^x\$
\$\displaystyle 1-x=e^x\$
I know that ln(e^x)=x so I thought of doing that
ln(1-x)=x
But I don't know what to do after that. Thanks for the assistance.
• Sep 19th 2010, 03:41 PM
Chris L T521
Quote:

Originally Posted by iamanoobatmath
Hello I am having difficulty solving this question. Any help would be greatly appreciated.
\$\displaystyle g(x)=3+x+e^x\$
The e is the mathematical constant not a variable

find \$\displaystyle g^{-1}(4)\$

So I must find a value of x where g(x)=4

this is my attempt:
\$\displaystyle 4=3+x+e^x\$
\$\displaystyle 1=x+e^x\$
\$\displaystyle 1-x=e^x\$
I know that ln(e^x)=x so I thought of doing that
ln(1-x)=x
But I don't know what to do after that. Thanks for the assistance.

At this stage, I can only complete it by observation:

If \$\displaystyle x=0\$, then \$\displaystyle 1-0=e^0\implies 1=1\$

So \$\displaystyle g^{-1}(4)=0\$
• Sep 19th 2010, 10:27 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
At this stage, I can only complete it by observation:

If \$\displaystyle x=0\$, then \$\displaystyle 1-0=e^0\implies 1=1\$

So \$\displaystyle g^{-1}(4)=0\$

Undoubtedly a solution by inspection was intended.