Question

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

Answer 1

see explanation

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

Answer 2

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[(3)/(4)x^4-(x^2)/(2)-x\right]_(-3)^(-2)`

`\\ \tt\Rrightarrow (3)/(4)(-2)^4-((-2)^2)/(2)-(-2)-\left((3)/(4)(-3)^4-((-3)^2)/(2)+3\right)`

`\\ \tt\Rrightarrow 12-2+2-((243)/(4)-(9)/(2)+2)`

`\\ \tt\Rrightarrow 12-((225)/(4)+2)`

`\\ \tt\Rrightarrow 12-(233)/(4)`

`\\ \tt\Rrightarrow (48-233)/(4)`

`\\ \tt\Rrightarrow (-185)/(4)`

`\\ \tt\Rrightarrow 46(approx)\\eq 0`

It has a zero in between-3 and -2