<s>[INST] f(x) is a real valued function,$ f(0) = 78$ and $f(x+2) -f(x) \leq 3 * 2^x$ and $f(x+6) -f(x) \geq 63 * 2^x$ $f(78) = k-1+ 2^k$. find k Let $f(x)$ be a real valued function such that $ f(0) = 78$ and $f(x+2) -f(x) \leq 3 * 2^x$ and $f(x+6) -f(x) \geq 63 \cdot 2^x$ $f(78) = k-1+ 2^k$. Find $k$.
I calculated $f(2)$, $f(4)$ and $f(6)$. I found the value of the function for $x=6$. What should be done further to solve the problem.
 [/INST] Using the first inequality three times we get that
\begin{align}
f(x+2) - f(x) &\leq 3\cdot 2^x\\
f(x+4) - f(x+2) &\leq 3\cdot 2^{x+2}\\
f(x+6) - f(x+4) &\leq 3\cdot 2^{x+4}
\end{align}
Adding all these inequalities yields
$$f(x+6)-f(x) \leq 3\cdot(1+2^2+2^4)\cdot 2^x = 63\cdot 2^x. $$
Together with the other given inequality this gives equality, i.e. 
$$f(x+6)-f(x) = 63\cdot 2^x,$$
or equivalently
$$f(x+6) = f(x) + 63\cdot 2^x. $$
You can now apply this multiple times starting with the known value when $x=0$ until you reach the value for $f(78) = f(13\cdot 6)$.
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