Im going through some notes for fractal geometry and the following exercise is stated:
"Let $f:[0,1]\rightarrow{}\mathbb{R}$ be a continuously differentiable function with $f(0)\neq{}f(1)$. Show that the box dimension of the image of $f$, namely $f([0,1])=\{f(x):0\le{}x\le{}1\}$ is $1$."
Firstly, my attempt:
Since f is differentiable on $[0,1]$ it is bounded on $[0,1]$ and hence has a maximum and minimum . The intermediate value theorem tells us that the image is also an interval, the result follows (simple calculation).
Secondly:
Is the question stated wrong? could it be in fact to calculate the box dimension of the graph $\{(x,f(x):0\le{x}\le{1})\}\subset\mathbb{R}^2$? Even if it's not an error, is the box dimension of this determinable? Is it dependant on the function?