Answered: Using a graphing device to solve the…

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter3: Polynomial And Rational Functions

Section3.7: Polynomial And Rational Inequalities

Problem 48E

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Question

Using a graphing device to solve the equality express your answer using interval notation, with the importance of the intervals rounded to two decimals. x^4-5x^3+18x>0

Expert Solution

Step 1

Given :

The inequality $x^{4} - 5 x^{3} + 18 x > 0$

To solve :

The inequality and express using interval notation.

Step 2

Given inequality is $x^{4} - 5 x^{3} + 18 x > 0$

To solve it for x, we take $x^{4} - 5 x^{3} + 18 x = 0$

$x^{4} - 5 x^{3} + 18 x = 0 x (x^{3} - 5 x^{2} + 18) = 0 x = 0 o r x^{3} - 5 x^{2} + 18 = 0$

Let $f (x) = x^{3} - 5 x^{2} + 18$ , then $f (3) = {(3)}^{3} - 5 {(3)}^{2} + 18 = 0$

$\begin{matrix} 1 - 5 0 18 \\ 0 3 - 6 - 18 \end{matrix} 1 - 2 - 6 0 3$

Therefore, the factors of

$x^{3} - 5 x^{2} + 18 = (x - 3) (x^{2} - 2 x - 6)$

Using quadratic formula to solve $x^{2} - 2 x - 6 = 0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ ,

where a = 1 , b = -2 and c = -6

$\begin{array}{rcl} x & = & \frac{2 \pm \sqrt{4 - 4 (1) (- 6)}}{2 \times 1} \\ = & \frac{2 \pm \sqrt{28}}{2} \\ = & \frac{2 \pm 2 \sqrt{7}}{2} \\ = & 1 \pm \sqrt{7} \end{array}$

Therefore, x = 0 , 3 , $1 + \sqrt{7} a n d 1 - \sqrt{7}$ .

x = 0 , - 1.65 , 3 and 3.65.

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