How do you find the inverse of #f(x)=(x+2)/(x-1)# and is it a function?

Question

How do you find the inverse of  #f(x)=(x+2)/(x-1)# and is it a function?

Chloe Alexander · Accepted Answer

#f^-1(x)=(x+2)/(x-1)# #f(x)=(x+2)/(x-1)# #y=(x+2)/(x-1)# #y*x-y=x+2# #y*x-x=y+2# #x(y-1)=(y+2)# #x=(y+2)/(y-1)# #"change x and y vice versa"# #y=(x+2)/(x-1)# #f^-1(x)=(x+2)/(x-1)# #"or simply:"# #f(x)=(a x+b)/(c x+d)# #f^-1 (x)=(-d x+b)/(c x+a)#

Serenity Jones · Answer

undefined To find the inverse of $ f(x) = \frac{x+2}{x-1} $, we first switch the roles of $ x $ and $ y $ and then solve for $ y $.

$$ x = \frac{y + 2}{y - 1} $$

Now, we solve for $ y $:

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

$$ xy - x = y + 2 $$

$$ xy - y = x + 2 $$

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

$$ y = \frac{x + 2}{x - 1} $$

So, the inverse function of $ f(x) = \frac{x+2}{x-1} $ is $ f^{-1}(x) = \frac{x + 2}{x - 1} $.

To determine if $ f(x) = \frac{x+2}{x-1} $ is a function, we need to check if for every $ x $ value, there is exactly one corresponding $ y $ value.

Since the function $ f(x) = \frac{x+2}{x-1} $ is defined for all real numbers except $ x = 1 $, it is a function. However, to be more precise, we should also check that the function is not undefined for any values in its domain. In this case, $ f(x) $ is undefined when $ x = 1 $, as it would result in division by zero. Therefore, we need to exclude $ x = 1 $ from the domain of the function. So, $ f(x) = \frac{x+2}{x-1} $ is a function defined for all real numbers except $ x = 1 $.