The point with a coordinate of (2,5) will be translated into y=-x line. The transformation for y=-x would be:
1. x'= -y
2. y'= -x
For x=2 and y=5 would be:
x'= -y
x'= -5
y'= -x
y'= -2
The translated coordinate would be: (-5, -2)
Start with #47. To find the critical values, you must differentiate this function. x times (4-x)^3 is a product, so use the product rule. The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x]
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of f '(x) = (4-x)^2 [-3x+4-x]. (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1. Thus, the "cv" are {4,1}.
Evaluate the given function at x: {4,1}. For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8. Thus, one of the extreme values is (1,8).
Answer:
5.673091 x 10^6
Step-by-step explanation:
