Chapter 1.1, Problem 33E

To determine

To find: find the domain of the given function.

Answer to Problem 33E

The domain of the function is, $\left(-\infty ,0\right)\cup \left(5,\infty \right)$

Explanation of Solution

Given information:

The given function is,

  $h\left(x\right)=\frac{1}{\sqrt[4]{x^2-5x}}$

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

  $h\left(x\right)=\frac{1}{\sqrt[4]{x^2-5x}}$

The fourth root a negative number is not allowed (as a real number)

That $h\left(x\right)$ is not defined when $x^2-5x\le 0$

The domain of $h$ consists of all value of $x$ such that

  $\begin{array}{c}{x}^2-5x\le 0\\ x^2-5x>0\\ x\left(x-5\right)>0\end{array}$

There are two conditions,

First condition,

  $x>0$

And,

  $\begin{array}{c}x-5>0\\ x>5\end{array}$

Second condition,

  $x<0$

And,

  $\begin{array}{c}x-5<0\\ x<5\end{array}$

Hence, the domain of $h$ is

  $\left\{x\in R|x\ne 0,x\ne 5\right\}$

Therefore the domain of $h$ is the interval,

  $\left(-\infty ,0\right)\cup \left(5,\infty \right)$