# Inverse Functions

• Sep 12th 2007, 04:44 AM
Tom G
Inverse Functions
I've got two functions [f(x) and g(x)] which I need to find the inverse of [f-(x) and g-(x)], can someone please explain the steps taken when finding the inverse of funtions.

1. f(x)= the square root of 'x' (i don't have a square root button on my keyboard)

2. g(x)= 2x+1

I would appreciate if someone could show each step they have taken to get the answer.
• Sep 12th 2007, 05:07 AM
topsquark
Quote:

Originally Posted by Tom G
I've got two functions [f(x) and g(x)] which I need to find the inverse of [f-(x) and g-(x)], can someone please explain the steps taken when finding the inverse of funtions.

1. f(x)= the square root of 'x' (i don't have a square root button on my keyboard)

2. g(x)= 2x+1

I would appreciate if someone could show each step they have taken to get the answer.

The general method is to take a function $\displaystyle y = f(x)$, reverse the roles of x and y ($\displaystyle x = f(y)$, then solve this equation for y ($\displaystyle y = g(x) = f^{-1}(x)$.)

When doing this you need to be careful about domains and ranges. (The range of the function is the domain of the inverse function and visa versa.)

For example:
1) $\displaystyle y = f(x) = \sqrt{x}$
Switch the roles of x and y: $\displaystyle x = \sqrt{y}$
Solve for y: $\displaystyle y = x^2$

Thus $\displaystyle f^{-1}(x) = x^2$.

Note, though, that since f(x) is only defined on $\displaystyle [0, \infty )$ that the inverse function is only defined on $\displaystyle [0, \infty )$ rather than $\displaystyle (-\infty, \infty )$.

2) $\displaystyle g(x) = 2x + 1$.
You do this one. I get $\displaystyle g^{-1}(x) = \frac{x - 1}{2}$.

-Dan
• Sep 12th 2007, 05:14 AM
red_dog
If $\displaystyle f:A\to B$ and $\displaystyle f(x)=y$ then $\displaystyle f^{-1}:B\to A$ and $\displaystyle f^{-1}(y)=x$.
So, to find $\displaystyle f^{-1}$ you have to find $\displaystyle y$ as a function of $\displaystyle x$.

1) $\displaystyle f:[0,\infty)\to[0,\infty),f(x)=\sqrt{x}$.
$\displaystyle \sqrt{x}=y$
Square both sides:
$\displaystyle x=y^2$ So, $\displaystyle f^{-1}(y)=y^2$
Now, change the letter for the variable:
$\displaystyle f^{-1}:[0,\infty)\to[0,\infty), \ f^{-1}(x)=x^2$

Now, can you solve 2)?
• Sep 12th 2007, 05:22 AM
Tom G
Thanks, I've now managed to do g(x) and got the same answer as topsquark.

I've got:

g(x)=2x+1

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

(x-1)/2=g-(x)