Prove f(x) ∈ O(x^3) Using Big-O Definition: Homework

[sta|ker]

Homework Statement

Let $f(x) = 2x^{3} + 3x\log{x}$, prove $f \in O(x^{3})$ using the Big-O Definition.

Homework Equations

Big-O definition:
$f(x) \in O(g(x))$ if $|f(x)| \leq C|g(x)|$ and $x \geq k$ where $C$ and $k$ are both positive integers.

The Attempt at a Solution

I basically set $C=4$ and $k=4$, then wrote it out:
$|2x^{3}+3x\log{x}| \leq 4|x^{3}|$ where $x \geq 4$

then using 4 for x:
$256 \geq 152$

According to the definition this proves it. It just seems to simple for an assignment (not the only problem on it, but still). Did I prove this correctly? Or do I completely not understand? I can prove it using limits, but she wants us to use the Big-O therom.

 

[Office_Shredder]

You have to prove it for EVERY x larger than 4, not just when x = 4. Also is your log base 2?

 

[LCKurtz]

Homework Statement

Let $f(x) = 2x^{3} + 3x\log{x}$, prove $f \in O(x^{3})$ using the Big-O Definition.

Homework Equations

Big-O definition:
$f(x) \in O(g(x))$ if $|f(x)| \leq C|g(x)|$ and $x \geq k$ where $C$ and $k$ are both positive integers.

The Attempt at a Solution

I basically set $C=4$ and $k=4$, then wrote it out:
$|2x^{3}+3x\log{x}| \leq 4|x^{3}|$ where $x \geq 4$

then using 4 for x:
$256 \geq 152$

According to the definition this proves it. It just seems to simple for an assignment (not the only problem on it, but still). Did I prove this correctly? Or do I completely not understand? I can prove it using limits, but she wants us to use the Big-O therom.

So saying $|2x^{3}+3x\log{x}| \leq 4|x^{3}|$ when $x>4$ you are observing that $3x\log x \le 2x^3$ when $x\ge 4$. I would think you would need to give an argument for that or at least explain why you already know that after checking $x=4$.

 

[sta|ker]

You have to prove it for EVERY x larger than 4, not just when x = 4. Also is your log base 2?

Yes, the log base is assumed to be 2. Sorry, I forgot about that.

So saying $|2x^{3}+3x\log{x}| \leq 4|x^{3}|$ when $x>4$ you are observing that $3x\log x \le 2x^3$ when $x\ge 4$. I would think you would need to give an argument for that or at least explain why you already know that after checking $x=4$.

Hmm, I guess I don't understand that technique then. I can do it using limits, but I guess I'll just have to review it until I get it.

 

[LCKurtz]

So saying $|2x^{3}+3x\log{x}| \leq 4|x^{3}|$ when $x>4$ you are observing that $3x\log x \le 2x^3$ when $x\ge 4$. I would think you would need to give an argument for that or at least explain why you already know that after checking $x=4$.

Hmm, I guess I don't understand that technique then. I can do it using limits, but I guess I'll just have to review it until I get it.

It's just an ordinary calculus question now. How could you argue that $3x\log_2 x \le 2x^3$ if $x>4$?