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Aloiza [94]
3 years ago
5

Please I need some help!

Mathematics
1 answer:
Schach [20]3 years ago
7 0

(f+g)(x) = 8x+2

(f-g)(x) = 4x-6

Step-by-step explanation:

Given

f(x) =6x-2\\g(x)=2x+4

we have to find

<u>a. (f+g)(x)</u>

(f+g)(x)=f(x)+g(x)\\=(6x-2)+(2x+4)\\=6x-2+2x+4\\=6x+2x-2+4\\=8x+2

<u>b. (f-g)(x)</u>

(f-g)(x)=f(x)-g(x)\\=(6x-2)-(2x+4)\\=6x-2-2x-4\\=6x-2x-2-4\\=4x-6

Hence,

(f+g)(x) = 8x+2

(f-g)(x) = 4x-6

Keywords: Functions, Operations on Functions

Learn more about functions at:

  • brainly.com/question/4021035
  • brainly.com/question/4034547

#LearnwithBrainly

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Answer:

1) W₁ is a subspace of Pₙ (R)

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4) W₃ is a subspace of Pₙ (R)

Step-by-step explanation:

Given that;

1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R

2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R

3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R

so

1)

let W₁ = { at² ║ a∈ R }

let ∝ = a₁t² and β = a₂t²  ∈W₁

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)

= c₁a₁t² + c²a₂t²

= (c₁a₁ + c²a₂)t² ∈ W₁

Therefore c₁∝ + c₂β ∈ W₁ for all ∝, β ∈ W₁  and scalars c₁, c₂

Thus, W₁ is a subspace of Pₙ (R)

2)

let W₂ = { t² + a ║ a∈ R }

the zero polynomial 0t² + 0 ∉ W₂

because the coefficient of t² is 0 but not 1

Thus W₂ is not a subspace of Pₙ (R)

3)

let W₃ = { at² + a ║ a∈ R }

let ∝ = a₁t² +a₁t  and β = a₂t² + a₂t ∈ W₃

let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t² +a₁t) + c₂(a₂t² + a₂t)

= c₁a₁t² +c₁a₁t + c₂a₂t² + c₂a₂t

= (c₁a₁ +c₂a₂)t² + (c₁a₁t + c₂a₂)t ∈ W₃

Therefore c₁∝ + c₂β ∈ W₃ for all ∝, β ∈ W₃ and scalars c₁, c₂

Thus, W₃ is a subspace of Pₙ (R)

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