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# Square Root Function

Sep 14, 2022

### 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.

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