question in french (use google please because i can't translate percisely)

[Amengnichi11]

question 2)a.

 

[Deleted member 4993]

question 2)a.
View attachment 22791

Is this a part of a test? The "stamp" on the upper right corner is "suspicious"!

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Amengnichi11]

yes it is i'am preparing for my test next week
the stamp shows the name of a college(المعهد النموذجي بصفاقس) in tunisia
im stuck in question 2)a. and this is my work

 

[HallsofIvy]

You are given the equation \(\displaystyle x^3- 9x- 12= 0\) and (2a) asks you to show that there is a solution in the interval (3.5, 3.6).

What is \(\displaystyle x^3- 9x- 12= 0\) when x= 3.5? What is \(\displaystyle x^3- 9x- 12= 0\) when x= 3.6? What does that tell you?

 

[Amengnichi11]

It's not x=3.5 it's alpha. And the exercice tells you to prove that when f(x) =0 it has a solution (alpha) that is contained between 3.5 and 3.6/he wants you to prove that there is a solution (alpha) and in another question he will ask for that numbers figure(question 2)d.)

 

[Dr.Peterson]

I think Halls means this:

What is \(\displaystyle x^3- 9x- 12\) when x= 3.5? What is \(\displaystyle x^3- 9x- 12\) when x= 3.6? What does that tell you?​

That is, what are the values of the function at the endpoints of the interval?

 

[Amengnichi11]

I think Halls means this:

What is \(\displaystyle x^3- 9x- 12\) when x= 3.5? What is \(\displaystyle x^3- 9x- 12\) when x= 3.6? What does that tell you?​

That is, what are the values of the function at the endpoints of the interval?

Thanks I got it?

 

[JeffM]

I am not going to try to answer in French.

First [MATH]f(x) = x^3 - 9x - 12 \ne x^3 - 9x - 19.[/MATH]
[MATH]\dfrac{(a^3 - 9a - 12) - (b^3 - 9b - 12)}{a - b} = \dfrac{a^3 - 9a - b^3 + 9b}{a - b} = [/MATH]
[MATH]\dfrac{(a^3 - b^3) - 9(a - b)}{(a - b)} = \dfrac{(a - b)(a^2 + ab + b^2) - 9(a - b)}{(a - b)} =[/MATH]
[MATH]a^2 + ab + b^2 - 9 \implies\\ \dfrac{f(a) - f(b)}{a - b} < 0 \iff a^2 + ab + b^2 < 9 \text { and}\\ \dfrac{f(a) - f(b)}{a - b} > 0 \iff a^2 + ab + b^2 > 9.[/MATH]That is correct as is your conclusion.

Although it is not clear to me how you arrived at it, also correct is your conclusion that f(x) is decreasing if and only if

[MATH]- \sqrt{3} < x < \sqrt{3}.[/MATH]
But that conclusion does not respond to question 2a.

f(x) is continuous at all x. (la fonction est continue a tout x?)

[MATH]\therefore a < b \text { et } f(a) * f(b) < 0 \implies \\ \exists \ c \text { telle que } a < c < b \text { et } f(c) = 0.[/MATH][MATH]f(3.5) = -0.625[/MATH]
[MATH]f(3.6) = 2.256[/MATH]
Donc?