Chapter 3.5, Problem 24AYU

To determine

To calculate: $f\left(x\right)=\sqrt{x-3x^2}$

Answer to Problem 24AYU

The domain of $f\left(x\right)=\sqrt{x-3x^2}$ is $0\le x\le \frac{1}{3}$

Explanation of Solution

Given Information:

  $f\left(x\right)=\sqrt{x-3x^2}$

Calculation:

The given function is:

  $f\left(x\right)=\sqrt{x-3x^2}$

The expression inside the square root cannot be negative is the domain condition.

So,

  $\begin{array}{l}x-3x^2\ge 0\\ x\left(1-3x\right)\ge 0\end{array}$

Here, $x\ge 0$

  $\begin{array}{l}\left(1-3x\right)\ge 0,\\ 1-3x\ge 0,\\ -3x\ge -1,\\ x\ge \frac{1}{3}\end{array}$

So, the domain of $f\left(x\right)=\sqrt{x-3x^2}$ is $0\le x\le \frac{1}{3}$