Finding an upperbound on $f(n)$ I am stumped trying to prove that there exists a real number $c$, such that $f(n)\leq cn^4$ for most natural numbers $n$. 
$$f(n) = \left\{
     \begin{array}{ll}
       10, &n=10\\
       3f\left(\left\lfloor \frac{2n}{5} \right\rfloor \right) + 6n^4,&n\geq 1
     \end{array}
   \right.$$
I have computed some values of $f(n)$ from $0-9$ and have chose my constants that seem to make $f(n)\leq cn^4$ true: $c=7$ and $n>4$. However, I don't know how to prove that $c$ is valid. Thanks for your time.

Use (strong) induction:
for $n$ big enough, such as $k< n\implies f(k) \le Ck^4$,
$$f(n) = 
3f\left(\left\lfloor \frac{2n}{5} \right\rfloor \right) + 6n^4
\le \left[ 3C \frac {2^4}{5^4} + 6\right]n^4
\le \left( 0.08 C + 6\right)n^4
\le Cn^4
$$
 as soon as
$$
 0.08 C + 6 \le C \Leftarrow C  \ge 7
$$
Then check that $7$ is convenient for the first values (until $n > \lfloor \frac{2n}{5} \rfloor $), and you are done.
