+130511 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 !!!
+130511 Here's (11)
Let the side of the square cut out = x
So, if you can imagine it....the area of the bottom of the box will be (20 -2x)(20-2x)....and the height of the box = x
So....the volume will be
V = x*(20 - 2x)^2 ..... let's just expand this out
V = x(4 x^2 - 80x + 400)
V = 4x^3 - 80x^2 + 400x
V' = 12x^2 - 160x + 400 divide through by 4
V' = 3x^2 - 40x + 100 let's see if this will factor (and set it to 0)
(3x -10)(x - 10) = 0
Reject the second factor...it would make a side length at the bottom = 0 !!
So we have
3x - 10 = 0
3x = 10
x = 10/3 .... this is the height
And the length of each side of the bottom = (20 - 2(10/3)) = (20 - 20/3) = (60/3 - 20/3) =40/3
And the volume is just...h x area of the base = (10/3)(40/3)^2 = about 592.6 cm^3
Here's a graph......https://www.desmos.com/calculator/zq8xlob32o
