has critical points wherever the partial derivatives vanish:
Then
- If
, then 
; critical point at (0, 0) - If
, then 
; critical point at (1, 1) - If
, then 
; critical point at (-1, -1)
has Hessian matrix
with determinant
- At (0, 0), the Hessian determinant is -16, which indicates a saddle point.
- At (1, 1), the determinant is 128, and
, which indicates a local minimum. - At (-1, -1), the determinant is again 128, and
, which indicates another local minimum.
Quadrant 2 because it goes to the left because the X is negative then up because the Y is positive.
The answer is -8
====================================================
Explanation:
There are two ways to get this answer
Method 1 will have us plug x = 0 into h(x) to get
h(x) = x^2 - 4
h(0) = 0^2 - 4
h(0) = 0 - 4
h(0) = -4
Then this output is plugged into g(x) to get
g(x) = 2x
g(-4) = 2*(-4)
g(-4) = -8 which is the answer
This works because (g o h)(0) is the same as g(h(0)). Note how h(0) is replaced with -4
So effectively g(h(0)) = -8 which is the same as (g o h)(0) = -8
-----------------------
The second method involves a bit algebra first
Start with the outer function g(x). Then replace every x with h(x). On the right side, we will replace h(x) with x^2-4 because h(x) = x^2-4
g(x) = 2x
g(x) = 2( x )
g(h(x)) = 2( h(x) ) ... replace every x with h(x)
g(h(x)) = 2( x^2-4 ) ... replace h(x) on the right side with x^2-4
g(h(x)) = 2x^2-8
(g o h)(x) = 2x^2-8
Now plug in x = 0
(g o h)(x) = 2x^2-8
(g o h)(0) = 2(0)^2-8
(g o h)(0) = 2(0)-8
(g o h)(0) = 0-8
(g o h)(0) = -8
Regardless of which method you use, the answer is -8
The last one, v = -2,000y + 20,000
