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3 years ago

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

Mathematics

2 answers:

geniusboy [140]3 years ago

6 0

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

elena-14-01-66 [18.8K]3 years ago

5 0

We need to integrate

3x^2-10-x+9=3x^3-x-1

$\\ \tt\Rrightarrow {\displaystyle{\int}_{-3}^{-2}}3x^2-x-1$

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

$\\ \tt\Rrightarrow \dfrac{3}{4}(-2)^4-\dfrac{(-2)^2}{2}-(-2)-\left(\dfrac{3}{4}(-3)^4-\dfrac{(-3)^2}{2}+3\right)$

$\\ \tt\Rrightarrow 12-2+2-(\dfrac{243}{4}-\dfrac{9}{2}+2)$

$\\ \tt\Rrightarrow 12-(\dfrac{225}{4}+2)$

$\\ \tt\Rrightarrow 12-\dfrac{233}{4}$

$\\ \tt\Rrightarrow \dfrac{48-233}{4}$

$\\ \tt\Rrightarrow \dfrac{-185}{4}$

$\\ \tt\Rrightarrow 46(approx)\neq 0$

It has a zero in between-3 and -2

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