The function
... y = 1/x
has derivative
... y' = -1/x²
which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.
Your function
... y = (x+1)/(x-3)
can be written as
... y = 1 +4/(x-3)
which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).
The critical point is x=3.
The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.
Answer:
a) True
b) False
c) False
d) False
e) True
Step-by-step explanation:
a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).
b) Suppose that 
is a basis of V. Considere the set 
where 
are scalars. The set has 5 vectors but 
because 
is not belong to A and is linear independent of 
c) Suppose that is a basis of V. Considere the set 
where 
are scalars. A has four nonzero vectors but isn't a basis because is a LD set.
d) Suppose that is a basis of V. Considere the set 
where are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.
e) Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.
It’s b i got that same question and got it right
Answer Jamies datab is greater than chuck,s data
Step-by-step explanation:
