# Fonction

• Oct 18th 2007, 07:22 AM
iceman1
Fonction
Hello Guys

I Need some Help in Fonctions
The function f is set to R:
F (x) = 3x 4x ^ 2 -1/2

1. Calculate f (-1), f (-1/2), f (1 / 2) and f (1) { i did this)
2. Proving that the equation f (x) = 0 admits three distinct solutions, ranging from
-1 And 1

3. Let x = cos (alfa)
Express cos3 (alfa), according to cos (alfa)
2. Determiner solutions of the equation f (x) = 0

my second question for x between 0 and 2pie, solve cos(4x)=sin(2x) {edit : i solved this No need to solve it again lol:D

and my last question :
Prove that the equation sinx -1/2 = 0
In admits } 0, 2 (pi) {a unique solution

Thanks In Advance :)
• Oct 18th 2007, 07:49 AM
earboth
Quote:

Originally Posted by iceman1
...
and my last question :
Prove that the equation sinx -1/2 = 0
In admits } 0, 2 (pi) {a unique solution

Hello,

if I understand your question right you are looking for the solutions of

$\sin(x) - \frac12 = 0~,\ x \in\ [0, 2\pi ]~\iff~\sin(x)=\frac12$

Then you get 2 (two) solutions: $x = \frac16 \cdot \pi~\vee~x = \frac56 \cdot \pi$ (see attachment)
• Oct 18th 2007, 06:07 PM
iceman1
Thanks Mate , any help with the First question?? i only solved the first part

Thanks
• Oct 19th 2007, 06:08 AM
topsquark
Quote:

Originally Posted by iceman1
I Need some Help in Fonctions
The function f is set to R:
F (x) = 3x 4x ^ 2 -1/2

1. Calculate f (-1), f (-1/2), f (1 / 2) and f (1) { i did this)
2. Proving that the equation f (x) = 0 admits three distinct solutions, ranging from
-1 And 1

Probably the reason no one has helped you with it is because the form is strange.

Is this
$F(x) = 12x^3 - 1/2$
or
$F(x) = 12x^2 - 1/2$
or something else?

Edit: to have 3 roots it's likely a cubic, but I would like verification of the form first.

-Dan
• Oct 20th 2007, 01:28 AM
earboth
Quote:

Originally Posted by iceman1
Hello Guys

I Need some Help in Fonctions
The function f is set to R:
F (x) = 3x 4x ^ 2 -1/2

1. Calculate f (-1), f (-1/2), f (1 / 2) and f (1) { i did this)
2. Proving that the equation f (x) = 0 admits three distinct solutions, ranging from
-1 And 1
...

Hello,

you mentioned 3 (three) distinct solutions. So I assume that your function reads:

$f(x) = 3x\left(4x^2-\frac12\right)$

You have to solve the equation f(x) = 0:

$3x\left(4x^2-\frac12\right) = 0$ . A product of two factors equals zero if one factor equals zero:

$3x\left(4x^2-\frac12\right) = 0~\iff~3x = 0~\vee~4x^2-\frac12 = 0$ Solve the equations for x. I've got: $x = 0 \vee x= -\frac14 \cdot \sqrt{2}~\vee~x= \frac14 \cdot \sqrt{2}$

These three solutions belong to [-1, 1]
• Oct 20th 2007, 08:10 AM
iceman1
Ah ok Thanks Guys :)