I have worked this over and over still think it's wrong.

find its inverse, domain, and range of f and f^-1
f(x) = 5/ (3x -7)

Can someone please help me with this???????

Too bad you didn't show us what your answers are. Maybe they are right, and it would save us the trouble of working it out (again?)

f is a rational function. s such, its domain is all reals except where the denominator is zero. That would be where x = 7/3

The range of f is all reals except 0. There is no number big enough to yield 0 when dividing it into 5.

f^-1(x) = (7x+5)/3x
Domain is all reals except x=0, and range is all reals except y = 7/3. As x gets large, the function approaches 7x/3x = 7/3, but does never equal it. Note that the horizontal asymptote is the same value as the vertical asymptote of f(x).

To find the inverse of a function, you need to switch the roles of x and y and solve for y. Let's start by doing that for function f(x) = 5/(3x - 7):

Step 1: Replace f(x) with y: y = 5/(3x - 7)
Step 2: Swap x and y: x = 5/(3y - 7)
Step 3: Solve the equation for y:

Multiply both sides by (3y - 7): (3y - 7)x = 5
Distribute on the left side: 3xy - 7x = 5
Move the constant term to the other side: 3xy = 5 + 7x
Divide by 3x to isolate y: y = (5 + 7x)/(3x)

This equation represents the inverse function f^(-1)(x). Now, let's determine the domain and range of both f(x) and f^(-1)(x).

For function f(x) = 5/(3x - 7):

Domain: The function f(x) is defined for all x values except when the denominator 3x - 7 is equal to zero. So to find the domain, solve the equation 3x - 7 = 0:

3x - 7 = 0
3x = 7
x = 7/3

Therefore, the domain of f(x) is all real numbers except x = 7/3, since that would make the denominator zero.

Range: To find the range, consider the behavior of the function as x approaches positive infinity and negative infinity. As x approaches positive infinity, the denominator 3x - 7 becomes extremely large, causing the value of f(x) to approach zero. As x approaches negative infinity, the denominator becomes extremely negative, leading f(x) to also approach zero. Therefore, the range of f(x) is all real numbers except zero.

Now, let's move on to the inverse function f^(-1)(x) = (5 + 7x)/(3x):

Domain: The inverse function f^(-1)(x) has the same domain as the original function f(x). Therefore, the domain of f^(-1)(x) is all real numbers except x = 7/3.

Range: The range of the inverse function is the set of all y-values for which f^(-1)(x) is defined. In this case, since f^(-1)(x) is a rational function, its range is all real numbers except when the denominator 3x becomes zero. Therefore, the range of f^(-1)(x) is all real numbers except x = 0.