I'm trying to understand the deep relations between the tangent line to the graph of a function $f$ at a given point $P$, and the derivative of $f$ at the same point.

Indeed, in many books the derivative is often defined as the slope of the "tangent", with only an intuitive definition of what a tangent line is. Then, once the concept of derivative is well assimilated, books define more precisely the tangent as... the line passing through $P = (a, f(a))$ with slope $f'(a)$ ! This seems a bit circular.

So some authors use an other approach, starting by giving a geometric definition of a tangent, and then showing the connection with derivatives. That's what one can read in the paper "What a tangent line is when it isn't a limit", by Irl Bivens. But I am missing something in the proof of Theorem 1, when the author writes:

it suffices to show that $|f(x)-L(x)|\leq (1/2)|L(x)-K(x)|$

where $f$ is a differentiable function, $L$ is the line of equation $L(x) = f'(a)(x-a)+f(a)$ and $K$ an other line, of equation $K(x)=p(x-a)+f(a)$ with $p \neq f'(a)$. So, basically, the goal is to show that $L$ is a better approximation of $f$ than any $K$.

But where does the right-hand side come from ? It looks like, locally, $f(x) = 1/2 (L(x)+K(x))$, but why?

  • $\begingroup$ Perhaps it would be helpful if you included a little more context here, at the very least describe what the symbols $f, L, K$ are are. $\endgroup$ – Travis Willse May 30 '15 at 14:36

Given that $$ |L(x)-K(x)|=|L(x)-f(x)+f(x)-K(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|\tag1, $$ if $$ |L(x)-K(x)|\geq2|f(x)-L(x)|\tag2 $$ then $$ 2|f(x)-L(x)|\leq|f(x)-L(x)|+|f(x)-K(x)|, $$ from which you obtain easily that $$ |f(x)-L(x)|\leq|f(x)-K(x)|\tag3. $$

  • $\begingroup$ Thank you, @enzotib; but I'm sorry, I don't understand where the first inequality comes from, neither the connection with my question? $\endgroup$ – Greg82 May 30 '15 at 16:05
  • $\begingroup$ @Greg82: $(1)$ is a simple application of the triangular inequality, $(2)$ is your inequality, $(3)$ is the inequality defining the tangent line, I show that $(1)+(2)\implies(3)$. $\endgroup$ – enzotib May 30 '15 at 16:08
  • $\begingroup$ @ enzotib: perfectly clear now, thanks a lot! $\endgroup$ – Greg82 May 30 '15 at 16:14

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