u20051503
Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.