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pop000 · Dec 9, 2010, 06:34 AM

1) how I find the cutting point with X axis of the function:6*sqrtIXI-X-8
I think is x=0 but I don't know how to show the way of it.

2)how I prove that the function:f(x)=cotx-tanx, are Continuous in(0;pi/2).
and there is no any Absolute minimum or Maximum.

thanks.

Unknown008 · Dec 9, 2010, 07:15 AM

1. No, the function should be equal to zero, not x. And you didn't make use of proper brackets...

Is that it?

$f(x) = 6\sqrt{|X|}-X-8$

2. $f(x) = \cot x - \tan x$

$f(x) = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}$

$f(x) = \frac{\cos^2x - \sin^2 x}{\cos x\sin x}$

$f(x) = \frac{\cos 2x}{\frac12 \sin 2x}$

$f(x) = \frac{2\cos 2x}{\sin 2x}$

$f(x) = 2\cot 2x$

You should know that cot(2x) is continuous between 0 and pi/2.

pop000 · Dec 9, 2010, 07:56 AM

'well yes I know in General that cotx are continuous between (0 and pi/2] and tanx are continuous between (-pi/2;pi/2).

But how I prove that there is no Absolute minimum or Maximum.
But there are local minimum or Maximum point

And really thanks about all you'r answers in every Subject almost :)

pop000 · Dec 9, 2010, 07:59 AM

by the way I mean x=4 is the point that the function are cutting the X axis
but how I show the way?

thanks

Unknown008 · Dec 9, 2010, 08:54 AM

Hm... I have seen this notation, but I'm afraid I won't be able to help you further for part 2 =/, sorry for that... and cot(x) is continuous for (0 and pi] while cot(2x) is continuous for (0 and pi/2], if I got that notation right.

1. $f(x) = 6\sqrt{|x|} - x - 8$

There are three solutions, one of which is indeed x = 4

$0 = 6\sqrt{|x|} - x - 8$

$6\sqrt{|x|} = x + 8$

$\sqrt{|x|} = \frac{x+8}{6}$

$|x| = \frac{x^2 + 16x + 64}{36}$

This gives:

$x = \frac{x^2 + 16x + 64}{36}$ or $x = \frac{-x^2 - 16x - 64}{36}$

$0 = x^2 - 20x + 64$ or $0 = x^2 + 52x + 64$

$x = \frac{20 \pm \sqrt{400 - 4(64)}}{2}$ or $x = \frac{-52 \pm \sqrt{2704 - 4(64)}}{2}$

$x = \frac{20 \pm 12}{2}$ or $x = \frac{-52 \pm 12\sqrt{17}}{2}$

$x = 10 \pm 6$ or $x = -26 \pm 6\sqrt{17}$

You get x = 4, x = 16, $x = -26 + 6\sqrt{17}$ and $x = -26 - 6\sqrt{17}$

Using these in the original equation, only x = 4, 16 and $-26 + 6\sqrt{17}$ are correct.

pop000 · Dec 9, 2010, 09:15 AM

Thanks now I Understand.