To algebraically determine the inverse of f(x), first replace f(x) with y. Then, switch x and y and solve for y.
Function
Inverse of the Function
Equation
f(x)=1/(x+1)^2
f^(- 1)(x)= ± 1/sqrt(x)-1
Domain
x≠1
x > 0
Range
y > 0
y≠ -1
Result: The inverse is not a function.
Practice makes perfect
We will begin by finding the inverse of f(x). First, we need to replace f(x) with y. From there, we will switch x and y and solve for y.
y=1/( x+1)^2 → x=1/( y+1)^2
The resulting equation will be the inverse of the given function.
Finally, to indicate that this is the inverse function of f(x), we replace y with f^(- 1)(x).
f^(- 1)(x)= ± 1/sqrt(x)-1
Now that we have found the inverse of f(x), we will determine the domain and range of f(x) and f^(-1).
Domain and Range of the Function
To determine the domain of the given function, remember that the denominator of a quotient cannot be 0. This means that we need to find the values of x that would make the denominator 0.
If x=-1, the denominator is 0. Therefore, the domain is the set of all real numbers except -1.
Domain:& x≠1
For the range, think about the fact that the square of a number is always positive.
f(x)=1/(x+1)^2 l← ← lpositive positive
Because the quotient of two positive numbers is always a positive number, the range is all values of y such that y>0.
Range:& y>0
Domain and Range of the Inverse
When we find the inverse of a function, we are basically exchanging the x and y values. Therefore, the range of f(x) becomes the domain of f^(- 1)(x), and the domain of f(x) becomes the range of f^(- 1)(x).
Domain: & x>0
Range:& y≠ 1
Is the Inverse a Function?
A function is a relation where each input is related to exactly one output. In our case, we can see that for each x in the domain of f^(- 1)(x), there are two values of y.
If x=a, then y=± 1/sqrt(a)-1.
Therefore, the inverse of the function is not a function.
