Letfbe the the vector space of all real functions. (the functions are "vectors" inf).a) (2 points) is the set of functions of the formf(x) =ax4+bx2+c, wherea, b,ccan be any real numbers, a subspace off? if not, explain why. b) (2 points) is the set of functions of the formf(x) =ax2+bx+ 1, whereaandbcan be any real numbers, a subspace off? if not, explain why. c) (4 points) find the basis for the function subspace containing the following functions{f(x) =a+bsin(x) +ccos(x)|a+b+c= 0; 2a−b=c}.

a) Yes, it is

b) No, it isnt

c) < sen(x) - cos (x) >

Step-by-step explanation:

A subset S from a R-vector space V is a subspace is:

a) Lets call A the set of all real valued functions of the form .

Given f ∈ A, , for certain real numbers a,b,c , and given r ∈ R. We have that

Since ra, rb and rc are real numbers, then r*f ∈ A.

Now lets suppose f1 are f2 are functions in A, with , . Then

Since a1+a2 , b1 +b2 and c1+c2 are all real numbers, then f1+f2 is an element of A.

Therefore A is a vector subspace of F.

b) Lets call B the subset of funtions of the form f(x) =ax²+bx+ 1. You can note that every single function of B have a 1, that means, each function f satisfies f(0) = 1. We can easily exploit this property to reach the conclussion that B is not a vector subspace of F.

For example, we can consider the function f(x) = x² + x + 1. Clearly f is an element of B because it satisfies the conditions. However if we multiply by the real number 2 we obtain that 2*f (x) = 2x² + 2x + 2, which is not an element of B because it doesnt have 1 as independent term. Therefore B is not a linear subspace.

c) Lets analyze what possible values a,b and c may have. Since a+b+c = 0, then c = -b-a. Replacing this value of c on the other equation we have that

2a-b = -b-a, or equivalently 3*a = 0

Thus, a = 0, and by replacing this value on the first equation we have that c = -b. As a result, (a,b,c) = (0,b,-b) = b * (0, 1, -1). Therefore, the combination (a,b,c)  must be a real multiple of (0,1,-1). Replacing by that value, we have that a basis for the subspace is < sen(x) - cos (x) >

I hope this helped you!

answer: r there any options?

step-by-step explanation:

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1