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 are. 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 about what I meant by “domain and range interchange” between the function and its inverse, see my previous lesson about this.


Examples of How to Find the Inverse of a Square Root Function

Example 1: Find the inverse function, if it exists. State its domain and range.

f(x) = sqrt of the quantity x+3 , x is greater than or equal to negative 3

Every time I encounter a square root function with a linear term inside the radical symbol, I always think of it as “half of a 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.

graph of f(x) = sqrt of the quantity x+3 where x is greater than or equal to -3 and y is greater than or equal to 0.

From this point, I will have to solve for the inverse algebraically by following the suggested steps. Basically, replace [latex]\color{red}f\left( x \right)[/latex] by [latex]\color{red}y[/latex], interchange [latex]x[/latex] and [latex]y[/latex] in the equation, solve for [latex]y[/latex] 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.

(f^-1)(x) = (x^2)-3; where x is greater than or equal to 0 and y is greater than or equal to -3.

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.

The graph of the original function, f(x) = sqrt of the quantity x+3, and its inverse, (f^-1)(x) = (x^2).

Can you see their symmetry along the line [latex]y = x[/latex]? See the green dashed line.


Example 2: Find the inverse function, if it exists. State its domain and range.

f(x) = negative sqrt of x minus 1; x is greater than or equal to 1

This function is the  “bottom half” of a parabola because the square root function is negative. That negative symbol is just [latex]-1[/latex] in disguise.

graph of f(x) = negative sqrt of x minus 1

In solving the equation, squaring both sides of the equation makes that [latex]-1[/latex] “disappear” since [latex]{\left( { – 1} \right)^2} = 1[/latex]. Its domain and range will be the swapped “version” of the original function.

(f^-1)(x) = (x^2)+1 where x is less than or equal to 0 and y is greater than or equal to 1

Example 3: Find the inverse function, if it exists. State its domain and range.

f(x) is equal to negative sqrt of x plus 5; x is greater than or equal to negative 5.

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 [latex]y-[/latex]axis while considering the x-values from the domain.

graph of f(x) is equal to negative sqrt of x plus 5

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.

(f^-1)(x) = (x^2)-5 where x is greater than or equal to 0 and y is greater than or equal to negative 5

Example 4: Find the inverse function, if it exists. State its domain and range.

f(x) is equal to the sqrt of 9 minus x squared, x is greater than or equal to negative 3 and less than or  equal to 0

This function is one-fourth (quarter) of a circle with radius [latex]3[/latex] 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 the time to graph it. Well, I hope that you realize the importance of having a visual aid to help determine that “elusive” range.

The graph of f(x) is equal to the sqrt of 9 minus x squared.

The presence of a squared term inside the radical symbol tells me that I will apply the square root operation on 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 on which one to pick as the correct inverse function. Remember that inverse function is unique therefore I can’t allow having 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.

(f^-1)(x) = negative sqrt of 9 minus x squared where x is greater than or equal to 0 and less than or equal to 3, AND y is greater than or equal to negative  3 and less than or equal to 0

Although they have the same domain, the range here is the “tie-breaker”! The range tells us that the inverse function has a minimum value of [latex]y = -3[/latex] and a maximum value of [latex]y = 0[/latex].

The positive square root case fails this condition since it has a minimum at [latex]y = 0[/latex] and maximum at [latex]y = 3[/latex]. The negative case must be the obvious choice, even with further analysis.


Example 5: Find the inverse function, if it exists. State its domain and range.

f(x) is equal to negative sqrt of 25 minus x squared, x is greater than or equal to 0 and less than or equal to 5

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 the horizontal axis. Notice that this is similar to Example 4. It is also one-fourth of a circle but with a radius of [latex]5[/latex]. The domain forces the quarter circle to stay in Quadrant IV.

graph of f(x) is equal to negative sqrt of 25 minus x squared

This is how we find its inverse algebraically.

Did you pick the correct inverse function out of the two possibilities? The answer is the case with the positive sign.

(f^-1)(x) = sqrt of 25 minus x squared where x is greater than or equal to negative 5 and less than or equal to 0 AND y is greater than or equal to 0 and less than or equal to 5

You may also be interested in these related math lessons or tutorials:

Inverse of a 2×2 Matrix

Inverse of Absolute Value Function

Inverse of Constant Function

Inverse of Exponential Function

Inverse of Linear Function

Inverse of Logarithmic Function

Inverse of Quadratic Function

Inverse of Rational Function