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.

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.

'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 :)

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

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.

Thanks now I Understand.