We're given f(x) equals the square root of the quantity two x minus one minus three. We want
to determine the domain and range of the given function and then find the inverse function.
Because our function contains a square root, in order for the function value to be real,
the number underneath the square root, or the radicand, which in this case is two x minus one,
can’t be negative. This means two x minus one must be greater than or equal to zero.
Since two x minus one must be greater than or equal to zero, this restriction will help us
find our domain. We just need to solve this for x. So, we'll add one to both sides and divide by two.
Thus, x has to be greater than or equal to one-half, which would be the domain of f(x).
If we wanted to use interval notation, this would be the interval from one-half to infinity,
closed on one-half, meaning it includes one-half.
To find the range of f(x), we want to determine all the possible y values of this function. The
value of the square root is always going to be greater than or equal to zero. So, the smallest
value this could be would be zero minus three, which would be negative three. All other function
values would be larger than negative three. So the range is y is greater than or equal to
negative three, or using interval notation, we’d have the interval from negative three to infinity.
Let’s go ahead and verify this by graphing the original function.
Here’s the graph of the original function. Notice if we project this function onto the x-axis,
the x-values start at x equals one-half,
and then from here, the x-values move to the right and approach positive infinity.
This verifies the domain.
Now to verify the range, we’ll project this function onto the y-axis. Notice how the y-values
start at y equals negative three, and then from here, the function does move right very quickly
but also moves up indefinitely. Therefore, the y-values approach
positive infinity. This graph does verify we have the domain and range correct for the function.
Remember, the domain of the function will become the
range of the inverse function, and the range of the function
will become the domain of the inverse function. So now let’s find the inverse of our function f(x).
Remember, inverse functions undo each other. This means if
function f has an input of x and an output of y,
this y becomes the input into the inverse function, which returns the original value of x.
The process for finding the inverse function is to write this in terms of x and y,
and then interchange the x and y variables. So we can write the given function as y
equals the square root of the quantity two x minus one minus three, which means the inverse function
would be the equation x equals the square root of the quantity two y minus one minus three.
Now we just need to solve this for y and replace y with inverse function notation. So,
the first step will be to add three to both sides of the equation. So,
we’ll have x plus three equals the square root of the quantity two y minus one.
To undo the square root, we’ll square both sides of the equation.
Let’s go ahead and leave this as the quantity x plus three squared on the left side.
On the right side, squaring the square root leaves us with the radicand of two y minus one.
Next step, we’ll add one to both sides of the equation.
That’ll give us the quantity x plus three squared plus one equals two y.
The last step instead of dividing by two, let’s multiply both sides by one-half.
So on the left side, we’d have one-half
times the quantity x plus three squared plus one equals on the right side we just have y.
Now we could multiply this out and try to simplify this but I'm going to go ahead
and leave it in this form. But we should replace y with inverse function notation. So, f inverse of x
is equal to one-half times the quantity x plus three squared plus one.
Now it’s important to remember the domain of this inverse function is restricted. It’s going to be
the range of the original function. So let’s go ahead and list that the domain of this function
is going to be when x is greater than or equal to negative three,
or the interval from negative three to infinity.
Again, remember the output of the original function becomes the input into the inverse
function. So the range of f is the domain of f inverse and the range of the inverse function
will be the domain of the original function. So our domain is going to be y
greater than or equal to one-half,
which using interval notation would be from one-half closed on one-half to positive infinity.
So here’s the domain of the original function, here’s our inverse function, and the domain and
range of the inverse function. The last step, I always like to verify this graphically. Remember,
if we graph the function and its inverse function on the same coordinate plane,
the two functions should be symmetrical across the line y equals x.
So here’s the graph of the original function, the square root function,
and here’s a graph of the quadratic function with the restricted domain.
And here’s the graph of y equals x.
Notice how these two functions are symmetrical across the line y equals x, which verifies they
are inverses of one another and verifies our work is correct. I hope you found this helpful.