Square Root Function

Function, Domain, and Codomain

Function

A function is an equation for which any x that can be put into the equation will produce exactly one output such as out of the equation.

It is represented as y=fx, where is an independent variable and y is a dependent variable.

Example:  y = x+2

Domain

A domain of a function is the set of inputs for which the function is defined. 

Codomain  

A codomain of a function is the set of possible output values. 

Square Root Function

The function f(x) = √x is the square root function.  

Graph of square root function:  

Properties of square root function:  

1. Domain = All values of xx such that x≥0.  
2. Range = f(x)≥0.  
3. For f(x)=√x, the x–intercept and y-intercept of the graph of the function are both 0. 

Note: The graph is increasing for all the values in the domain of f.  

Example:  

Sketch the graph p(x)=-√x and find the intercept, domain, and range of the function.  

Solution:  

Function = √𝒇𝒙=−𝒙

Domain = All values of x such that x≥0.  

Range = f(x)≤0 .  

For f(x)=√x, the x–intercept and y–intercept of the graph of the function are both 0. 

Translation of the square root function

Example: The graph of g(x)=√x+3 compared to the graph of f(x) = √x.

Solution:  

The graph of g(x) = √x+3 is the vertical translation of f(x) = √x.

The domain for both functions is x≥0.  

The range for the function f is y≥0, so the range for the function g is y≥3.  

Example:  

The graph of g(x)=√x+3 compared to the graph of f(x)=√x.

 Solution:  

The graph of g(x)=√x+3 is the horizontal translation of f(x)=√x.

The domain for function f is x≥0, so the domain for function g is x≥−3.  

The range for both functions is y≥0. 

Rate of Change of Square Root Function   

Example:  

For the function f(x)=√x, how does the average rate of change from x = 0 to x=0.3 compare to the average rate of change from x = 0.3 to x = 0.6?  

Solution:  

Step 1:  

Evaluate the function for the x–x-values that correspond to the endpoints of each interval.  

f(0) = √0 =0

f(0.3) = √0.3 ≈ 0.548

f(0.6) = √0.6 ≈ 0.775

Step 2:  

Find the average rate of change over each interval.  

From x=0 to x=0.3: 

f(0.3)−f(0)/0.3−0 ≈ 0.548−0/0.3−0 = 0.548/0.3 ≈1.83

From x=0.3 to x=0.6:  

f(0.6)−f(0.3)/0.6−0.3 ≈ 0.775−0.548/0.6−0.3 =0.227/0.3 ≈ 0.757

The average rate of change over the interval 0≤x≤0.3 is greater than the average rate of change over the interval 0.3≤x≤0.6.

Example:  

Two plans are being considered to determine the speed of a theme park ride with a circular wall that spins. Plan A is represented by the function in the graph shown. The ride shown in the photo is an example of plan B. If the ride has a radius of 5 m, which plan will result in greater speed for the ride? 

Solution:   

Plan A:  

The graph of plan A shows that the corresponding speed at a radius of 4 m is about 6 m/s.  

Plan B:  

Evaluate f(r)=√2r for r=5.  

f(5) = 2√5 ≈ 4.47

The ride using plan B has a speed of about 4.47 m/s when the radius is 5 m.  

Conclusion:  

With a radius of 4 m, the speed of the ride using plan A is 6 m/s.  

The speed of the ride using plan B is about 4.47 m/s.  

So, the ride using plan A has a greater speed for a radius of 4 m. 

Exercise

1. Compare the graph of g(x) = √x+4 to the graph of f(x) = √x.
2. Compare the graph of p(x) = √x-2  to the graph of f(x) = √x.
3. Find the average rate of change of f(x) = √3x ; 0≤x≤5.  

Concept Summary