Radical Functions

One of our basic functions is the square root function, it comes from the family of radical functions.
f(x)=xf(x) = \sqrt{x}
Radical functions contain radicals and are important for modeling a variety of situations.

Example 1
The following are examples of radical functions
  • f(x)=2x+13f(x) = 2\sqrt{x +1} - 3
  • g(x)=x+5g(x) = -\sqrt{x} + 5
  • h(x)=(x3)h(x) = \sqrt{-(x-3)}

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Domain and Range

For radical functions the index has an impact on the values that are in the domain and range of the function.
If the index is:
Even
Then the radicand must be positive. The radical will only output values like zero or greater.
As a result there will often be restrictions in our domain and range.

Odd
The radicand can be positive or negative. It will produce values that are both positive and negative.
As a result there are often no restrictions in our domain.

Example 2
Write the domain and range of the following

1. f(x)=4x5f(x) = 4\sqrt{x - 5}

The radicand will be positive when
x50x5\begin{aligned} x - 5 &\geq 0 \\ x &\geq 5 \end{aligned}
  • domain is [5,)[5, \infty)
  • range is [0,)[0, \infty)

2. g(x)=6x3+3g(x) = \sqrt[3]{6 - x} + 3

The index is odd so we have that
  • domain is (,)(-\infty, \infty)
  • range is (,)(-\infty, \infty)

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Distance Formula

One common application of radical function comes from modeling the distance between two points (x1,y1)(x_1, y_1)and (x2,y2)(x_2, y_2)
The distance formula is given by
d=(x2x1)2+(y2y1)2\boxed{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}
Example 3
Find the distance between the points (2, 3) and (x, 5)

d=(x2)2+(53)2d=(x2)2+4\begin{aligned} d &= \sqrt{(x - 2)^2 + (5-3)^2} \\ d &= \sqrt{(x-2)^2 + 4} \end{aligned}
This formula represents all the possible distances between the point (2,3)(2,3)
and the horizontal line y=5y=5

Example: Radical Functions

The surface area of a cone is given by the formula

SA=πrr2+h2\boxed{SA = \pi r \sqrt{r^2 + h^2}}
Where rr and hh are the radius of the base of the cone, and the height of the cone respectively.



The solid rocket boosters on the space shuttle form the shape of a cone.

1. If the diameter of the solid rocket boosters are about 3.7 meters. Create a function that describes the amount of metal needed for the surface of the cone as a function of the height.

S(h)=π(3.7)(3.7)2+h2=3.7π13.69+h2\begin{aligned} S(h) &= \pi (3.7) \sqrt{(3.7)^2 + h^2} \\ &= 3.7\pi\sqrt{13.69 + h^2} \end{aligned}

2. What is the domain of this function?

Although the equation has h2h^2 which ensures that the radicand will never be negative, since it represents height we can only consider non-negative values.

Domain [0,)[0, \infty)
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