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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. 33 (page 209)

Find the bounds to the zeros of each polynomial function. Use the bounds to obtain a complete graph of f.
$f (x) = 2 x^{3} + x^{2} - 1$

Short Answer

Expert verified

Every zero of the polynomial function will lie between $1$ and $- 1$ .

The graph of the function is as follows,

Step by step solution

01

Step 1. Given Information

We are given a polynomial function,

$f (x) = 2 x^{3} + x^{2} - 1$

We need to find the bounds to the zeros of the function and using it we need to obtain its graph.

02

Step 2. Concept used

Let f denote a polynomial function whose leading coefficient is $1$ .

$f (x) = x^{n} + a_{n - 1} x^{n - 1} + . . . + a_{1} x + a_{0}$

A bound M on the real zeros of f is the smaller of the two numbers,

$M a x \{1, |a_{0}| + |a_{1}| + . . . + |a_{n - 1}|\}, 1 + M a x \{|a_{0}|, |a_{1}|, . . ., |a_{n - 1}|\}$

where Max means 鈥渃hoose the largest entry in .鈥�

03

Step 3. Make the leading coefficient 1

The leading coefficient of the given polynomial is $2$ . So divide the function by $2$ , we get

$f (x) = x^{3} + \frac{1}{2} x^{2} - \frac{1}{2}$

On comparing with standard form we get,

$a_{2} = \frac{1}{2} a_{1} = 0 a_{0} = - \frac{1}{2}$

04

Step 4. Find the two numbers

Using the formula the two numbers are found as,

$Max \{1, |a_{0}| + |a_{1}| + 鈥� + |a_{n - 1}|\} = Max \{1, |- \frac{1}{2}| + | 0 | + |\frac{1}{2}|\} = Max {1, 1} = 1$

And,

$1 + Max \{|a_{0}|, |a_{1}|, 鈥�, |a_{n - 1}|\} = 1 + Max \{|- \frac{1}{2}|, 0, \frac{1}{2}\} = 1 + \frac{1}{2} = \frac{2 + 1}{2} = \frac{3}{2}$

05

Step 5. Find the bounds

Among the two numbers $1$ and $\frac{3}{2}$ , $1$ is the smallest.

Therefore, $1$ is the bound.

Every real zero off lies between $1$ and $- 1$ .

06

Step 6. Graphing the function

Using the bound the graph of the function is as follows,

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Most popular questions from this chapter

Find the domain of the rational function.

$G (x) = \frac{x - 3}{x^{2} + 1}$

Solve each inequality algebraically.

$\frac{x - 3}{x + 1} > 0$

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$f (x) = x^{3} + 2 x^{2} - 5 x - 6$

Solve the inequality algebraically

$x^{3} - 2 x^{2} - 3 x > 0$

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$f (x) = 4 x^{4} - 15 x^{3} - 8 x^{2} + 15 x + 4$

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