Given
subject to the constraint
Let
.
The gradient vectors of
and
are:
and
By Lagrange's theorem, there is a number
, such that
It can be seen that
has local extreme values at the given region.
For number 2 it's would be 12.8
Sorry I don't have time for number 3
The table of values of the function y = 2x^2 is:
x 0 1 2 3 4
y 0 2 8 18 32
<h3>How to complete the table?</h3>
The equation of the function is given as:
y = 2x^2
When x= 0, we have
y = 2 * 0^2 = 0
When x= 1, we have
y = 2 * 1^2 = 2
When x= 2, we have
y = 2 * 2^2 = 8
When x= 3, we have
y = 2 * 3^2 = 18
When x= 4, we have
y = 2 * 4^2 = 32
So, the table of values is:
x 0 1 2 3 4
y 0 2 8 18 32
Read more about functions and tables at:
brainly.com/question/15602982
#SPJ1
Answer:
x² - 64
Step-by-step explanation:
Given
(x + 8)(x - 8)
Each term in the second factor is multiplied by each term in the first factor, that is
x(x - 8) + 8(x - 8) ← distribute both parenthesis
= x² - 8x + 8x - 64 ← collect like terms
= x² - 64
the difference of 5 times the cube of x cubed and the quotient of 4 times x and 3
5*x^3 - (4x/3)
