Find the domain of the function $f(x) = \frac{\sqrt{x-2}}{\sqrt{5}} + \sqrt{(x - 1)(x + 2)(x + 4)}.$

Guest Sep 1, 2023

#1**+1 **

When we are dealing with square roots, the domain of the square root is restricted to when the radicand, also known as the argument of the square root, must be greater than or equal to zero. There are two radicands that contain variables, and we need to find the values where all of them produce values greater than or equal to zero.

The simplest of these to determine the domain is the term \(\sqrt{x - 2}\). This one is relatively straightforward.

\(x - 2 \geq 0 \\ x \geq 2\)

Now, we deal with the harder term, \(\sqrt{(x - 1)(x + 2)(x + 4)}\). Once again, the radicand must be greater than or equal to zero.

\((x - 1)(x + 2)(x + 4) \geq 0 \\ x = 1 \text{ or } x = -2 \text{ or } x = -4 \)

By using the Zero Product Property, we found x-values, x = 1, x = -2, and x = -4 that make the left-hand side of the inequality equal to zero. Now, we investigate the sign of the product to the left and right of these boundary x-values where the value changes from positive to negative or otherwise.

\(x < -4\) | \(-4 < x < -2\) | \(-2 < x < 1\) | \(x > 1\) | |

Sign of \((x - 1)(x + 2)(x + 4)\) | - | + | - | + |

Now that we have created the sign chart, we now know when the radicand is greater than or equal to zero. This means that we know that the restriction on the domain values are \(x \geq 2 \text{ and } (-4 \leq x \leq -2 \text{ or } x \geq 1)\). Put more simply, this just simplifies to \(x \geq 2\).

Written in interval notation, the domain of \(f(x)\) is \([2, \infty)\)

The3Mathketeers Sep 3, 2023

#1**+1 **

Best Answer

When we are dealing with square roots, the domain of the square root is restricted to when the radicand, also known as the argument of the square root, must be greater than or equal to zero. There are two radicands that contain variables, and we need to find the values where all of them produce values greater than or equal to zero.

The simplest of these to determine the domain is the term \(\sqrt{x - 2}\). This one is relatively straightforward.

\(x - 2 \geq 0 \\ x \geq 2\)

Now, we deal with the harder term, \(\sqrt{(x - 1)(x + 2)(x + 4)}\). Once again, the radicand must be greater than or equal to zero.

\((x - 1)(x + 2)(x + 4) \geq 0 \\ x = 1 \text{ or } x = -2 \text{ or } x = -4 \)

By using the Zero Product Property, we found x-values, x = 1, x = -2, and x = -4 that make the left-hand side of the inequality equal to zero. Now, we investigate the sign of the product to the left and right of these boundary x-values where the value changes from positive to negative or otherwise.

\(x < -4\) | \(-4 < x < -2\) | \(-2 < x < 1\) | \(x > 1\) | |

Sign of \((x - 1)(x + 2)(x + 4)\) | - | + | - | + |

Now that we have created the sign chart, we now know when the radicand is greater than or equal to zero. This means that we know that the restriction on the domain values are \(x \geq 2 \text{ and } (-4 \leq x \leq -2 \text{ or } x \geq 1)\). Put more simply, this just simplifies to \(x \geq 2\).

Written in interval notation, the domain of \(f(x)\) is \([2, \infty)\)

The3Mathketeers Sep 3, 2023