Chapter 1.1, Problem 31E

To determine

To find: find the domain of the given function.

Answer to Problem 31E

The domain of the function is $\left[-\infty ,\infty \right]$ ,

Explanation of Solution

Given information:

The given function is,

  $f\left(t\right)=\sqrt[3]{2t-1}$

Concept used:

To represent the function:

A function $f$ is a rule that assigns to each element $x$ in a set $D$ exactly one element called $f\left(x\right)$ in a set E. The set $D$ and $E$ are sets of real numbers. The set $D$ is called the domain of the function.

If $f$ is a function with domain $D$ .

Then the set of ordered pairs are.

  $\left\{\left(x,f\left(x\right)\right)|x\in D\right\}$

The given function is

  $f\left(t\right)=\sqrt[3]{2t-1}$

We take $f\left(t\right)=r$ to get the values of $t$ of the function so calculated as,

  $\begin{array}{c}\sqrt[3]{2t-1}=r\\ 2t-1=r^3\\ 2t=r^3+1\\ t=\frac{r^3+1}{2}\end{array}$

The cube root an odd number is not defined (as a real number)

The domain of $f$ consists of all value of $t$

  $\begin{array}{l}{r}^3>0,\\ r^3<0,\\ r^3=0\end{array}$

Now,

  $\begin{array}{l}{r}^3\in \infty ,\\ r^3\in -\infty \end{array}$

Then,

  $t\in \pm \infty$

Thus, the domain of $f$ is

  $\left\{t\in R|t\ne -\infty ,t\ne \infty \right\}$

Hence, this function is defined for all real number.

Therefore the domain of $f\left(t\right)$ is the interval,

  $\left[-\infty ,\infty \right]$