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A function $f:I\subseteq \mathbb{R} \rightarrow \mathbb{R}$ that is differentiable at $x_0 \in I$ obeys the following equality for all $h\in (I-x_0)$ (i.e. for all $h\in \mathbb{R}$ such that $x_0+h \in I $): $$ f(x_0+h)=f(x_0)+hf'(x_0)+hR(h).$$

This is nothing else than saying that the $f$ can be linearly approximated at $x_0$. My question is: How can I visualize this ? [For more details, read on]

Look at the following sketch I quickly made (sorry for the ugly mathfonts): enter image description here

In green it's sketched how the $f(x_0)$, the $hf'(x_0)$ and the $hR(h)$ should add up to give $ f(x_0+h)$.

My (more specific) question is: How on earth can I represent the number $f'(x_0)$, that geometrically can be made sense of via an angle of measure $\arctan(f'(x_0))$, as a segment which is the geometric interpretation of $hf'(x_0)$ (how one deals geometrically with multiplication with $h$ is already in a general matter described in this question of mine, so that not a problem) ? Geometrically we are thus adding a segment and a "scaled angle"!

I believe an equivalent question (correct me if I'm wrong) is: How can I determine the position of the blue point $S$ only with geometric means - which would mean only from the angle of measure $\arctan(f'(x_0))$ ?

[An even more complicated - but more of a side question, since it builds on the other more pressing question - how does the error term $R(h)$ (which involves $f'(x_0)$ in it's algebraic description) translate to its corresponding segment ?]

Do notice that the fact that proof of the equivalence of the fact that $f$ is differentiable at $x_0$ (and this definition can be fully illustrated geometrically) and that $f$ posses a best linear approximation at $x_0$ (whose geometric representation is the content of this question) does not give any clue how to represent these quantities geometrically since it consists only of algebraic manipulations and limiting processes. So somehow the "ability to represent algebraic quanities geometrically" is lost during the process of algebraic manipulations and taking limits.

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Add a horizontal and a vertical line segment to the line segment joining $(x_0,f(x_0))$ to $(x_0+h,f(x_0)+h f'(x))$ to make a right-angled triangle. Then $hf'(x_0)$ is the length of the vertical side. The triangle which is similar to this with horizontal length $1$ has vertical length $f'(x_0)$. Does this answer your question? – user108903 Dec 21 '12 at 16:36
@user108903 Yes! Perfect example of me being trapped in my analysis box, not being able to think outside it in terms of synthetic geometry! – temo Dec 22 '12 at 10:48
@user108903 Please consider posting your comment as an answer, since it indeed answered the question. – ˈjuː.zɚ79365 Jun 27 '13 at 9:15

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