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 $f\left(x\right)=\sqrt[n]{nr\left(x\right)}$, where $r\left(x\right)$ is an algebraic expression. For example $f\left(x\right)=\sqrt{x+1}$. When $n$ is odd number, $r\left(x\right)$ can be any real number. When $n$ is even, $r\left(x\right)$ has to be nonnegative, that is $r\left(x\right)\ge 0$ so that $f\left(x\right)$ is a real number.

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

Solution.

Since the index is $2$ which is even, the function has real outputs only if the radicand $x+1\ge 0$. Solve the inequality, we get $x\ge -1$. 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 $2\sqrt{L}=1$. $$\begin{array}{cc}2\sqrt{L}& =1\\ \sqrt{L}& =\frac{1}{2}\\ L& ={\left(\frac{1}{2}\right)}^2\\ L& =\frac{1}{4}.\end{array}$$ 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.

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