Find the only f=vt that has f(2t)=4f(t) I'm just starting calculus 1 and I don't know how to solve this.  Can someone please help?
The problem below involves linear functions $f(t) = vt + C$. Find the constants v and C.
Find the only $f=vt$ that has $f(2t)=4f(t)$.  show that every $f=\frac{1}{2}at^2$ has this property.
Assuming $v$ is constant: $f(2t) = 4f(t)$ implies $v(2t) = 4(vt)$ and therefore $2vt = 0$.
This yields either $t$ or $v$ must be $0$, but since the equation has to hold for all $t$ this means $v$ must be $0$.
Now let $f(t) = \frac{1}{2}at^2$. $f(2t) = \frac{1}{2}a(2t)^2 = \frac{1}{2}a(4t^2) = 4(\frac{1}{2}at^2) = 4f(t)$, regardless of $a$.
Hope this helps!
Let $t=2^s$.
Then
$$f(2t)=4f(t)$$ is
$$f(2\cdot2^s)=f(2^{s+1})=4f(2^s)$$ or $$g(s+1)=4g(s).$$
As no other information is given, $g$ is an arbitrary function in a unit range, and its value in other ranges are derived from the above recurrence, as
$$g(s+n)=4^ng(s)$$
or
$$f(2^nt)=4^nf(t)$$ where $f$ is arbitrary in, say, $[1,2)$.

$vt$ cannot work, because $v\,2^nt\ne 4^nv\,t.$
$at^2$ does, because $a\,(2^nt)^2=4^na\,t^2.$