Melody

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I got this from the insert drop down box - special characters - it is on the third row :)

Well put it back TR !!

Can I have some please ?

That is a good question Dragonlance.

I have laid it out a little better now so hopefully it is a little clearer    

$$f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\$$

The signs on the top belong with f(+1)

and the signs on the bottom belong with f(-1)

$$\\f(x)=\frac{Log[1 + 3*Sin[x]^2]}{2}\\\\
f(\pi/2)=\frac{Log[1 + 3*Sin[\pi/2]^2]}{2}\\\\
$now is this meant to be $[sin(pi/2)]^2\;\;or\;\;\;sin[(pi/2)^2] \;\;??\\\\
$I will assume $\;\;[sin(pi/2)]^2\\\\
f(\pi/2)=\frac{Log[1 + 3*[sin(pi/2)]^2]}{2}\\\\
f(\pi/2)=\frac{Log[1 + 3*1]}{2}\\\\
f(\pi/2)=\frac{Log[2]}{2}\\\\
$ now you can do it on a calc$\\\\\\
f(0)=\frac{Log[1 + 3*[sin(0)]^2]]}{2}\\\\
f(0)=\frac{Log[1 + 0]}{2}\\\\
f(0)=0\\\\$$

Yes, CPhill's way is better :)

$$\\(x^3-3x^2)-(2x-6)=0\\
x^3-3x^2-2x+6=0$$

Now use remainder theorem to find a root

$$\\let\\
f(x)=x^3-3x^2-2x+6\\
f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\
f(3)=27-27-6+6=0\qquad therefore\;\;x=3 \;\;$is a root$$$

 (this has been edited a little)

I used polynomial division to determine that

$$\\(x^3-3x^2-2x+6)\div(x-3)=(x^2-2)\\\\
so\\\\
x^3-3x^2-2x+6=0\\\\
(x-3)(x-\sqrt2)(x+\sqrt2)=0\\\\
x=3\quad or\quad x=\sqrt2\quad or \quad x=-\sqrt2$$

minus temperatures are very cold.  Positive ones are hot.

If you take away cold air from a room then the room gets hotter.   

So        - - = +

3 - - 4 = 3+4 = 7

Here you go, I love thissite.

LOL  Dragonlance   

I don't thinks TR had you all in the bag at the same time.  It was the same bag but  the others had each been dumped out of it already.   

"joker puts Dragonlance in the same kidnapping bad Rosala and 7up have been in"

See, it is past tense, you get the bag all to yourself!  Wow, what luxury.  :D