Determine whether the following are Vector Spaces

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SUMMARY

The discussion centers on determining whether specific sets are vector spaces. The set of real polynomials divisible by x² + x + 1 is confirmed as a vector space due to closure under addition and scalar multiplication. The second set, differentiable functions whose derivative equals 3x², requires clarification, as not all functions that differentiate to 3x² are included. The third set consists of functions f in F[0,2] such that x ≥ |f(x)| for 0 ≤ x ≤ 2, with examples provided to illustrate the concept.

PREREQUISITES
  • Understanding of vector space properties
  • Knowledge of polynomial functions and their divisibility
  • Familiarity with differentiation and derivatives
  • Concept of function bounds and inequalities
NEXT STEPS
  • Study vector space axioms and properties
  • Learn about polynomial divisibility and its implications
  • Explore differentiation techniques and their applications
  • Investigate function behavior under constraints and inequalities
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Mathematics students, educators, and anyone studying linear algebra or calculus who seeks to deepen their understanding of vector spaces and function properties.

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Homework Statement



a) The set of real polynomials of [itex]x[/itex] divisible by [itex]x^2 + x + 1[/itex];
b) The set of differentiable functions of [itex]x[/itex] on [itex][0,1][/itex] whose derivative is [itex]3x^2[/itex]
c) all [itex]f \in F[0,2][/itex] such that [itex]x \geq |f(x)|[/itex] for [itex]0 \leq x \leq 2[/itex]

The Attempt at a Solution



a) Yes, it's a vector space, proven with addition and scalar multiplication

b) I don't really understand what the question is saying, can someone explain to me? A function that differentiates to [itex]3x^2[/itex] is [itex]x^3[/itex]. Now what?

c) Same goes for this part, not sure what the question is saying

Thanks in advanced.
 
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(b) That is not all the functions that differentiate to ##3x^2##
(c) Try telling us what you think it's saying so we can see where the confusion lies.
 
f(x)= (2/3)x has the property that x> |f(x)|.

What about 5f(x)?
 

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