Check where the first-order partial derivatives vanish to find any critical points within the given region:
The Hessian for this function is
with 
, so unfortunately the second partial derivative test fails. However, if we take 
we see that 
for different values of 
; if we take 
we see 
takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.
Now check for points along the boundary. We can parameterize the boundary by
with 
. This turns into a univariate function 
:
At these critical points, we get
We only care about 3 of these results.
So to recap, we found that attains
- a maximum value of 4096 at the points (0, 8) and (0, -8), and
- a minimum value of -1024 at the point (-8, 0).
Answer:
2)
(2x -5) + (x) + (2x - 3) = 32.
5x-8 = 32.
ii) 5x-8=32
5x= 32+8
5x = 40
x= 40/5
x = 8
iii) 2x -5
2 × 8 -5 = 11.
2x-3
2×8-3
16-3 = 13
x =8
sides of the triangle= 8, 11, 13.
Relative extrema occur where the derivative is zero (at least for your polynomial function).
So taking the derivative we get
<span>20<span>x3</span>−3<span>x2</span>+6=0
</span><span>
This is a 3rd degree equation, now if we are working with complex numbers this equation is guaranteed to have 3 solutions by the fundamental theorem of algebra. But the number of real roots are 1 which can be found out by using Descartes' rule of signs. So the maximum number of relative extrema are 1.</span>
To solve for x, we will begin by using the distributive property
6x - 2(x + 4) = 12
6x - 2x - 8 = 12
(always remember to sort your negatives!)
Add like terms
4x - 8 = 12
Add 8 to both sides
4x = 20
Divide both sides by 4
x = 5
Answer:
B)law of detachment
Step-by-step explanation:
5u56i566i56
