Answer:
The fifth degree Taylor polynomial of g(x) is increasing around x=-1
Step-by-step explanation:
Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

and when you do its derivative:
1) the constant term renders zero,
2) the following term (term of order 1, the linear term) renders:
since the derivative of (x+1) is one,
3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero
Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is:
as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1
Answer:
3324333
Step-by-step explanation:
Answer:
2 b is you answer
Step-by-step explanation:
The ancestors to the equation is c=20
do you need to include the wiggle infront of the first bracket? if not;
(x+1) ÷ [(x^2+2) x (2x-3dx)]
x^2 x 2x = 2x^3
x^2 x -3dx = -3dx^3
2 x 2x = 4x
2 x -3dx = - 6dx
i cant find a way to make it equal 0 so i think the answer is just
x+1 over 2x^3 - 3dx^3 + 4x - 6dx as a fraction