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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 4: Q. 70 (page 242)

What is the domain of the function $f (x) = \sqrt{x^{3} - 3 x^{2}} ?$

Short Answer

Expert verified

The domain of the function is a set of all real numbers greater than or equal to $3 .$

Step by step solution

01

Step 1. Given Information

The given function is $f (x) = \sqrt{x^{3} - 3 x^{2}}$

We have to find the domain.

02

Step 2. Finding the domain

As the given function is a square root function.

To find the domain first, find the real zeros of f(x).

So, $\sqrt{x^{3} - 3 x^{2}} = 0 x^{3} - 3 x^{2} = 0 x^{2} (x - 3) = 0 x^{2} = 0 a n d x - 3 = 0 x = 0 x = 3$

03

Step 3. Determining intervals

Now, by using the real zeros the intervals we get are

$(- 鈭�, 0), (0, 3), (3, 鈭�) .$

04

Step 4. Check the points

Let $x = - 2$ and substitute in the given function

$f (- 2) = \sqrt{{(- 2)}^{3} - 3 {(- 2)}^{2}} = \sqrt{- 8 - 12} = \sqrt{- 20}$

So, it is not true.

Let $x = 2$ and substitute in the given function

$f (2) = \sqrt{{(2)}^{3} - 3 {(2)}^{2}} = \sqrt{8 - 12} = \sqrt{- 4}$

So, it is not true.

Let $x = 4$ and substitute in the given function

$f (4) = \sqrt{{(4)}^{3} - 3 {(4)}^{2}} = \sqrt{64 - 48} = \sqrt{16} = 4$

So, it is true.

Therefore, the domain of the function is all real numbers greater than or equal to $3 .$

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Most popular questions from this chapter

In Problems 63鈥�72, find the real solutions of each equation.

$2 x^{3} - 3 x^{2} - 3 x - 5 = 0$

In Problems 63鈥�72, find the real solutions of each equation.

$3 x^{3} - x^{2} - 15 x + 5 = 0$

Find a rational function that might have the given graph.

Find the real zeros of f. Use the real zeros to factor f. $f (x) = x^{4} + x^{3} - 3 x^{2} - x + 2$

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.

$F (x) = \frac{x^{4} - 16}{x^{2} - 2 x}$

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