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If we have a function $f(x) = 6x^4 - 2x^3 + 5$ and that function is $mathcalO(x^4)$. Does that mean that it will also be $Omega(x^4)$ and consequently $Theta(x^4)$?

Presumably you're talking about $x to infty$. Then it is true that $f(x) = mathcal O(x^4)$ and that $f(x) = Omega(x^4)$, so $f(x) = Theta(x^4)$. However, it's not always true that a function that is $mathcal O(x^4)$ is also $Omega(x^4)$. For example, $x^3 = mathcal O(x^4)$, but not $Omega(x^4)$.

In terms of $xtoinfty$, you have that a function $f$ is $Theta(g)$ if it is $mathcal O(g)$ and $Omega(g)$ per definition, so you definitely have the inverse of your statement.

Coincidentally, the function you described is $Theta(x^4)$ anyway. However, it is not always the case that $finmathcal O(g)$ implies $finTheta(g)$.

You can heuristically read $Theta$ as "equals", $mathcal O$ as less or equal than and $Omega$ as greater or equal then.

As an example, you have e.g. that $1inmathcal O(x^4)$ but not $1inOmega(x^4)$, thus $1notinTheta(x^4)$.

For your specific example, yes. In general: No.

Informally put: Let $f: mathbbR^+ mapsto mathbbR^+$ and $g: mathbbZ mapsto mathbbR^+$ be two positive real-valued functions whose domain is the set of positive reals.

1. Then $g$ is $O(f)$ iff there are constant $c>0$ such that $g(x) le c times f(x)$ for all $x$.




2. Then $g$ is $Omega(f)$ iff there is a constant $c'>0$ such that $g(x) > c' times f(x)$ for all $x$.




3. Then $g$ is $theta(f)$ iff BOTH $g$ is $O(f)$ AND $g$ is $Omega(f()$.




4. If $g$ is $O(f)$ but not $Omega(f)$ then $g$ is $o(f)$,




5. If $g$ is $Omega(f)$ but not $O(f)$ then $g$ is $omega(f)$.

Now letting $f$ be as in your question and $g = x^4$ indeed $g$ is both $Omega(f)$ and $g$ is $O(f)$, so $g$ is indeed $theta(f)$. But this follows becasue $f$ and $g$ are both polynomials of the same degree.