The domain of $f(x)$ is $\mathbb{R}$. For all $x \in \mathbb{R}$, $f(x+1)=2f(x)$. For $x \in (0,1]$, $f(x)=x(x-1)$. Find the least value of $k$ so that $f(x) \geq -\frac{3}{4}$ for all $x \in (-\infty, k]$.
I tried manipulating first $f(x+1)=2f(x)$, which gives me $f(x)=2^m(x-m)$, where $m$ is any integer less than $x$ (or it could mean the floor function). I also observe that it is always true for all $x \in (0,1]$.
I tried also using Autograph and look for pattern algebraically. Well, the thing is, I was really stucked on the sound pedagogy in solving this problem without the help of algebra.
If you could share some insights and ideas or strategies in solving this, it would be a great help.
I assume you mean "greatest $k$", since the problem has no solution if we're looking for a "least $k$" (because for any $k$ that works, so does $k - 1$).
The strategy involves noting that the minimal value of $f$ on the interval $[n, n+1]$ for $n$ integral is $-\frac{1}{4} 2^n$. This can be shown first by establishing the result on $[0, 1]$ (using some calculus or plain old completing the square) and then extending it to $[n, n+1]$ for positive $n$ by induction, then for negative $n$ by induction.
This means that the value in question must (if it exists) be in the interval $[2, 3]$. Over this interval, we find $f(x) = 4 (x - 2)(x - 3)$. Note that on the interval $[2, 5/2]$, $f$ is strictly decreasing. Further, note that $f(2) = 0$ and $f(5/2) = -1$. Find the $k$ in $(2, 5/2)$ such that $f(k) = -3/4$ (this $k$ exists by the intermediate value theorem). It turns out that this $k$ is $9/4$.
Now suppose that $x \leq k$. Case 1: $x \in [n, n+1]$ for some $n \leq 1$. Then $f(x) \geq -\frac{1}{4}2^n \geq -\frac{1}{2}$. Case 2: $x \in [2, k]$. Since $f$ is decreasing over $[2, k]$, we have $f(x) \geq f(k) = -3/4$. In either case, $x \leq k$ implies $f(x) \geq -3/4$.
Suppose we have some $k'$ where for all $x \leq k'$, we have $f(x) \geq -3/4$. Then suppose for sake of contradiction that $k' > k$. Then we can take some $x \in (k, 3] \cap [k, k')$. Since $f$ is strictly decreasing on $[k, 3]$, we would have $f(x) < f(k) = -3/4$. But $x < k'$ and thus $f(x) \geq -3/4$; contradiction. Then $k' \leq k$.
Thus, we have shown that $k = 9/4$ is the greatest $k$ such that for all $x \leq k$, $f(x) \geq -\frac{3}{4}$.