About the domain

I see you have made a start on this one

You have the first derivative

2 /  [x^3 √(2/x^2 - 1/x^4)]^(1/2)    which we can rewrite as

2 / [x^3 √(2x^-2 - x^-4)] =

(2x^-3)(2x^-2 - x^-4)^(-1/2)

Note that this derivative is never 0 because there s nothing in the numerator that would make it = 0

The second deivative is

(-6x^-4)(2x^-2 - x^-4)^(-1/2) + (2x^-3)(-1/2)(2x^-2 - x^-4)^(-3/2)(-4x^-3 + 4x^-5) =

(-6x^-4)(2x^-2 - x^-4)^(-1/2) + (2x^-3)(2x^-2 - x^-4)^(-3/2)(2x^-3 - 2x^-5)   factor this

[(2x^-4)(2x^-2 - x^-4)^(-3/2)] [ -3(2x^-2 - x^-4) + (x)(2x^-3 - 2x^-5)] =

[(2x^-4)(2x^-2 - x^-4)^(-3/2)][-6x^-2 + 3x^-4 + 2x^-2 - 2x^-4] =

[(2x^-4)(2x^-2 - x^-4)^(-3/2)] [x^-4 - 4x^-2]

We need to look at this diffcult one with several separate graphs (this is the only way I see this....)

Here's the graph of  sin^-1(1 - 1/x^2)...............https://www.desmos.com/calculator/lgofrknb1d

Notice that this graph has the same domain as the derivative........and notice that it is concave down at all points where it is defined

Now......let's look  at the second derivative in pieces

Here's the graph of the first two functions .........https://www.desmos.com/calculator/bqzeqzo4x4

And notice that these graphs are positive for all values in the domain

But.....let's look at the graph of the last function.........https://www.desmos.com/calculator/ekkinenwjq

Notice that this graph is negative for all values in the domain.......and putting this together with the first two functions, which are both positive, we see that the curve is concave downward at all points because...... (first two functions positive) x (last function negative) = negative

And there aren't any critical points to evaluate...the curve is always concave down, but there aren't any "max's".....the only "critical" points are ±1√/2......but at these points...the graph has a vertical tangent....

I think this is correct.......let me know if you have questions.....I did the best I could with this one !!!

Notice xvxvxv, that x cannot be 0  in the denominator

For the function iside the root....it must be > 0

Here's it's graph...........https://www.desmos.com/calculator/mfb0scvufp

The two points where this = 0 are at ±(1/√2)......then note, that on [-1/√2, 1/√2], this function is ≤ 0

So....the domain appears to be  (-∞, -1/√2) U ( 1/√2, ∞ ) 

But why zero is in the interval 

this is the question number 16

Oops, sorry...ignore that statement.....I got sidetracked !!!   I meant to say that 0 was in the excluded interval!!!

The domain is (-∞, -1/√2) U ( 1/√2, ∞ )

And 0 obviously isn't in that interval !!!!!

could you please solve the question from the bigining and show ,e the maximum and the minimum values 

I'm really need the answer for 16 

I want the answer for 16 

any one ? 

Question 16

sin(y) must have values between -1 and +1 so x must be less than or equal to -(√2)/2 or greater than or equal to (√2)/2. So the critical values of x are ±(√2)/2.

$$\frac{dy}{dx}=\frac{2}{x^3 \sqrt{1-(x^{-2}-1)^2}}$$

At x = ±(√2)/2, dy/dx = ∞

So these critical points are neither maxima nor minima

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I have a question, is the domain of sin inverse ( -1, 1)  

I've edited my original answer above, as I rushed the calculation of the critical points and got them wrong at first!  I've now corrected them.

You get them by solving 1-1/x² = -1

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Thank you Alan and Chris