Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$

Question

How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?

What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$ for values $x>1$. $C = 3/2, k = 1$

Is this right?

Answer 1

Hint: The degree of the numerator is $4$, while the degree of the denominator is $3$. The degree of $O(x)$ is $4-3=1$.

Answer 2

We have

$$ f(x)= \frac{x^4 + x^2 +1 }{x^3 +1} \leq C x, $$

for $x \geq \sqrt{\frac{C}{3(C-1)}}$ and $C>1$.

Answer 3

$$\dfrac{x^4+x^2+1}{x^3+1} = \dfrac{(x^2+x+1)(x^2-x+1)}{(x+1)(x^2-x+1)} = \dfrac{x^2+x+1}{x+1} = x + \dfrac1{x+1}$$ Now you should be able to conclude what you want.