can you translate it to engish, i might help!
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
R = { (x,y): 3x-y=0 }
The condition is 3x=y so that's not going to be any of these things.
R is reflexive if (x,x)∈R for all x. Let's check.
3x - y = 3x - x = 2x ≠ 0 necessarily. NOT REFLEXIVE
R is symmetric if (x,y)∈R → (y,x)∈R. Let's check.
(x,y)∈R so
3x-y = 0
y = 3x
Is (y,x)∈R. That would be true if 3y-x=0
3y - x = 3(3x) - x = 8x ≠ 0 necessarily NOT SYMMETRIC
R is transitive if (x,y)∈R and (y,z)∈R → (x,z)∈R. Let's check.
3x-y = 0 so y=3x
3y-z = 0 so z=3y = 9x
3x - z = 3x - 9x = -6x ≠ 0 necessarily NOT TRANSITIVE
Answer:
Step-by-step explanation:
Please use " ^ " to indicate exponentiation: f(x) = x^2 and g(x) = (3x)^2.
g(x) can be rewritten as 9x^2.
The graph of g(x) is only one ninth as wide as that of f(x).
Next time please share the possible answer choices. Thank you.
Answer:
true!
Step-by-step explanation:
because if we havent got lines, colors , and textures
how could shapes be created?