Finding the Inverse Function of a Square Root Function
To find the inverse of a square root function, it is crucial to sketch or graph the given problem first to clearly identify what the domain and range. Since I will utilize the domain and range of the original function to describe the domain and range of the inverse function by interchanging them. If you need additional information what I meant by “domain and range interchange” between the function and its inverse, see my previous lesson about this.
Example 1: Find the inverse function of , if it exists. State its domain and range.
Every time I encounter a square root function with a linear term inside the radical symbol, I always think of it as “half of parabola” that is drawn sideways. Since this is the positive case of the square root function, I am sure that its range will become increasingly more positive, in plain words, skyrocket to positive infinity.
This particular square root function has this graph, with its domain and range identified.
From this point, I will have to solve for the inverse algebraically by following the suggested steps. Basically, replace f(x) by y, interchange x and y in the equation, solve for y which soon will be replaced by the appropriate inverse notation, and finally state the domain and range.
Remember to use the techniques in solving radical equations to solve for the inverse. Squaring or raising to the second power the square root term should eliminate the radical. However, you must do it to both sides of the equation to keep it balanced.
Make sure that you verify the domain and range of the inverse function from the original function. They must be “opposite of each other”.
Placing the graphs of the original function and its inverse in one coordinate axis…
Can you see their symmetry along the line y = x? See green dashed line.
Example 2: Find the inverse function of , if it exists. State its domain and range.
This function is the “bottom half” of a parabola because the square root function is negative. That negative symbol is just −1 in disguise.
In solving the equation, squaring both sides of the equation makes that −1 “disappear” since (−1)2 = 1. Its domain and range will be the swapped “version” of the original function.
Example 3: Find the inverse function of , if it exists. State its domain and range.
This is the graph of the original function showing both its domain and range.
Determining the range is usually a challenge. The best approach to find it is to use the graph of the given function with its domain. Analyze how the function behaves along the “y-axis” while considering the x-values from the domain.
Here are the steps to solve or find the inverse of the given square root function.
As you can see, it’s really simple. Make sure that you do it carefully to prevent any unnecessary algebraic errors.
Example 4: Find the inverse function of , if it exists. State its domain and range.
This function is one-fourth (quarter) of a circle with radius 3 located at Quadrant II. Another way of seeing it, this is half of the semi-circle located above the horizontal axis.
I know that it will pass the horizontal line test because no horizontal line will intersect it more than once. This is a good candidate to have an inverse function.
Again, I am able to easily describe the range because I have spent time to graph it. Well, I hope that you realize the importance of having a visual aid to help determine that “elusive” range.
The presence of a squared term inside the radical symbol tells me that I will apply square root operation in both sides of the equation to find the inverse. By doing so, I will have a plus or minus case. This is a situation where I will make a decision which one to pick as the correct inverse function. Remember that inverse function is unique therefore I can’t allow to have two answers.
How will I decide which one to choose? The key is to consider the domain and range of the original function. I will swap them to get the domain and range of the inverse function. Use this information to match which of the two candidate functions satisfy the required conditions.
Although they have the same domain, the range here is the “tie-breaker”! The range tells us that the inverse function has minimum value of y = − 3 and maximum value of y = 0.
The positive square root case fails this condition since it has a minimum at y = 0 and maximum at y = 3. The negative case must be the obvious choice, even with further analysis.
Example 5: Find the inverse function of , if it exists. State its domain and range.
It’s helpful to see the graph of the original function because we can easily figure out both its domain and range.
The negative sign of the square root function implies that it is found below horizontal axis. Notice that this is similar to Example 4 . It is also one-fourth of a circle but with radius of 5. The domain forces the quarter circle to stay in Quadrant IV.
This is how we find its inverse algebraically.
Did you pick the correct inverse function out of the two possible ones? The answer is the case with the positive sign.