# Topic 15 Radical Functions

## 15.1 Speed of a Tsunami

A tsunami is generally referred to is a series of waves on the ocean caused by earthquakes or other events that cause sudden displacements of large volumes of water. In ideal situation, the velocity \(v\) of a wave at where the water depth is \(d\) meters is approximately \[ v=\sqrt{9.8d}. \] The wave will slow down when closer to the coast but will be higher.

Suppose a tsunami was caused by earthquake somewhere 10000 meters away the coast of California. The depth of the water where the tsunami was generated is 5000 meter.

- What’s the initial speed of the tsunami?
- What’s the speed of the tsunami at where the water depth is 2000.
- Suppose the speed wouldn’t decrease, how long it takes the tsunami reach the coast?

## 15.2 The Domain of a Radical Function

A ** radical function** \(f\) is defined by an equation \(f(x)=\sqrt[n]{r(x)}\), where \(r(x)\) is an algebraic expression. For example \(f(x)=\sqrt{x+1}\). When \(n\) is odd number, \(r(x)\) can be any real number. When \(n\) is even, \(r(x)\) has to be nonnegative, that is \(r(x)\geq 0\) so that \(f(x)\) is a real number.

**Example 15.1 **Find the domain of the function \(f(x)=\sqrt{x+1}\).

*Solution*.

Since the index is \(2\) which is even, the function has real outputs only if the radicand \(x+1\geq 0\). Solve the inequality, we get \(x\geq -1\). In interval notation, the domain is \[ [-1,\infty). \]

**Example 15.2 **For a pendulum clock, the period T of the pendulum is approximately modeled by the following function of the length L of the pendulum:
\[T=2\sqrt{L}\]
where \(L\) and \(T\) are measured in meters and seconds respectively.

- If the length of the pendulum is 4 meters, what is the period?
- If the period of a pendulum clock is 1 second, how long should be the pendulum?

*Solution*.- Because the length of the pendulum is 4 meters, that is \(L=4\). Then the period is \(T=2\cdot\sqrt{4}=4\) seconds.
- Because the period of the pendulum is 1 second, that is \(T=1\). Then the length \(L\) is the solution of the equations \(2\sqrt{L}=1\). \[\begin{aligned} 2\sqrt{L}&=1\\ \sqrt{L}&=\frac12\\ L&=\left(\frac12\right)^2\\ L&=\frac14. \end{aligned} \] So the length of the pendulum should be 1/4 in so that the period is 1 second.

## 15.3 Practice

**Problem 15.1 **Find the domain of each function. Write in interval notation.

- \(f(x)=1-\frac{2x}{x+3}\)
- \(f(x)=\frac{x-2}{x^2-4}\)
- \(f(x)=\sqrt{1-x^2}\)
- \(f(x)=-\sqrt{\frac{1}{x-5}}\)