Determine for which positive integer values of n, 3n^3+2≤n^4 and prove your claim by mathematical induction. I was able to determine n must be greater or equal to 4. But I can't seem to get through the inductive step.

Let $f(x)=3x^3+2-x^4, x>0$

$f'(x)=9x^2-4x^3$

Find the critical number(s)

$f'(x)=0=>9x^2-4x^3=0$

$x^2(9-4x)=0$

$x_1=0, x_2=2.25$

If $x<2.25, f'(x)>0, f(x)$ increases.

If $x>2.25, f'(x)<0, f(x)$ decreases.

$f(1)=3+2-1=4>0$

$f(2)=24+2-16=10>0$

$f(3)=81+2-81=2>0$

Therefore $3n^3+2\leq n^4$ for $n\geq4$.

Let $P(n)$ be the proposition that $3n^3+2\leq n^4, n\geq4.$

Basis Step

$P(4)$ is true, because $3(4)^3+2=194<256=(4)^4.$

Inductive Step

We assume that $3k^3+2\leq k^4.$ Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$3(k+1)^3+2\leq (k+1)^4$

We have

$3(k+1)^3+2=3k^3+9k^2+9k+3+2$

$\leq k^4+(4k)k^2+(6k)k+(4k+1)=(k+1)^4, k\geq4$

Hence $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for integers $n\geq4$. That is, we have proven that

$3n^3+2\leq n^4, n\geq4.$