f(x)=4/x-1
To do so we first need to replace f(x) with y. From there, we switch x and y and solve for y.
y=4/x-1 → x=4/y-1
The resulting equation will be the inverse of the given function.
Finally, to indicate that this is the inverse function of f(x) we will replace y with f^(- 1)(x).
f^(- 1)(x) = 4/x+1
b
Let's begin with finding f(f^(-1)(x)). To do so we will substitute x= f^(-1)(x) into our function, f(x)= 4x-1. Let's remember that we found f^(-1)(x)= 4x+1 in Part A.
x= f^(-1)(x) ⇒ x= 4/x+1
Therefore, we will substitute x= 4x+1 into the function f(x).
We can conclude that the outputs for both of the composite functions are x.
f(f^(-1)(x)) ⇔ x ⇔ f^(-1)(f(x))
Moreover, since the inputs and also the outputs are the same for f(f^(-1)(x)) and f^(-1)(f(x)), they are called identity functions.
c
We will examine the domain and range of the functions f and f^(-1). Let's begin with f.
f(x) = 4/x-1
Note that f(x) is a rational function which is only undefined where the denominator is zero.
x- 1=0 ⇒ x= 1
Since its denominator is 0 at x= 1, its domain is all real numbers except 1.
Now, let's recall that the range is the set of all outputs of a function. Notice that the division of a number by another number never gives 0. Therefore, the range of the function is the set of all real numbers except 0.
Domain: {x | All real numbers except 1 }
Range: {y | All real numbers except 0 }
Now, let's examine the function f^(-1)(x).
f^(-1) = 4/x+1
Since 4x is also a rational function, it is undefined where x= 0. Therefore, its domain is all real numbers except 0. For the range, since 4x never equals 0 we cannot get 1 as a result. Therefore, the range of the inverse function is the set of real numbers except 1.
Domain: {x | All real numbers except 0 }
Range: {y | All real numbers except 1 }
Notice that 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).
