Show that any continuous function defined for all real $x$ and satisfying the equation $\mathrm{f}\left(x\right) = \mathrm{f}\left(2x + 1\right)$ for all $x$ must be a constant function.
I do not know how to approach this question. Please guide anyone.
Suppose $f(a)\neq f(b)$ for some pair of points $a,b$.
If we use the equation with $2x+1=a$, this is, with $x=\frac{a-1}{2}$ we get a new point $\frac{a-1}{2}$ that takes the same value $f(a)=f((a-1)/2)$. We can now use again the equation with $2x+1=\frac{a-1}{2}$ and produce another point $x=\frac{\frac{a-1}{2}-1}{2}=\frac{a}{4}-\frac{1}{2}-\frac{1}{4}$ at which the function takes the same value $f(a)$. Continuing in this way we get a sequence of points $a,\frac{a-1}{2}, \frac{a}{4}-\frac{1}{2}-\frac{1}{4},...,\frac{a}{2^n}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2^n},...$ at which the function takes the same value $f(a)$. Notice that the sequence of points tends to $-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2^n}-...=-1$.
Now if we had two points $a,b$ such that $f(a)\neq f(b)$ we get two sequences of points tending to $-1$ such that the function takes two different values on them. Therefore the function wouldn't have a limit at $-1$. It wouldn't be continuous there.