u20051503
Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Show that the polynomial function f(x)=3x^3-10-x+9 has a real zero between -3 and -2

Sagot :

We need to integrate

  • 3x^2-10-x+9=3x^3-x-1

[tex]\\ \tt\Rrightarrow {\displaystyle{\int}_{-3}^{-2}}3x^2-x-1[/tex]

[tex]\\ \tt\Rrightarrow \left[\dfrac{3}{4}x^4-\dfrac{x^2}{2}-x\right]_{-3}^{-2}[/tex]

[tex]\\ \tt\Rrightarrow \dfrac{3}{4}(-2)^4-\dfrac{(-2)^2}{2}-(-2)-\left(\dfrac{3}{4}(-3)^4-\dfrac{(-3)^2}{2}+3\right)[/tex]

[tex]\\ \tt\Rrightarrow 12-2+2-(\dfrac{243}{4}-\dfrac{9}{2}+2)[/tex]

[tex]\\ \tt\Rrightarrow 12-(\dfrac{225}{4}+2)[/tex]

[tex]\\ \tt\Rrightarrow 12-\dfrac{233}{4}[/tex]

[tex]\\ \tt\Rrightarrow \dfrac{48-233}{4}[/tex]

[tex]\\ \tt\Rrightarrow \dfrac{-185}{4}[/tex]

[tex]\\ \tt\Rrightarrow 46(approx)\neq 0[/tex]

It has a zero in between-3 and -2

Answer:

see explanation

Step-by-step explanation:

Evaluate f(x) for x = - 3 and - 2

f(- 3) = 3(- 3)² - 10(- 3) + 9 = 3(- 27) + 30 + 9 = - 81 + 39 = - 42

Then (- 3, - 42 ) is below the x- axis

f(- 2) = 3(- 2)³ - 10(- 2) + 9 = 3(- 8) + 20 + 9 = - 24 + 29 = 5

Then (- 2, 5 ) is above the x- axis

Since f(x) is below the x- axis at x = - 3 and above the x- axis at x = - 2

Then it must cross the x- axis between x = - 3 and x = - 2

Indicating there is a real zero between - 3 and - 2

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.