If the traditional way to define the tangent line to a curve $f(x)$ through the point say $(a , f(a))$ is: ( the tangent line through the point $(a ,f(a))$ is the line that passes through this point with a slope that is equal to the value of the derivative at that point). From this definition why it follows that in many cases the tangent line touches the curve only once locally?

For example take the parabola $f(x)= x^2 $ the tangent line at the point $(1,1)$ has the equation $g(x)= 2x-1$. $g$ touches $f$ only at $(1,1)$. Why this is true not only for this special case, but for most curves at most points? I have an intuitive feeling for why this is true, so, is there a proof that the tangent line (from the above definition) will have this property (it touches the curve only once locally) for certain curves like: a circle or a polynomial of degree more than 1?

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    $\begingroup$ The "other definition" is odd. Is the line $x=0$ tangent to the parabola $y=x^2$ at $(0,0)$? $\endgroup$
    – saulspatz
    May 24, 2019 at 17:32
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    $\begingroup$ Your second definition of tangent line is incorrect. Every horizontal line meets the curve $y=\ln x$ in exactly one point, but none of them is a tangent to that curve. In fact, except for exceptionally "wobbly" curves, every line that passes through a curve will meet it in just a single point in a sufficiently small neighborhood of that point. $\endgroup$
    – MPW
    May 24, 2019 at 17:33
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    $\begingroup$ No, your question isn't clear at all to me. How can you expect to show that a correct definition and an incorrect one are equivalent? $\endgroup$
    – saulspatz
    May 24, 2019 at 17:36
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    $\begingroup$ It's not about "touching" the graph, nor how many times it touches the graph. It's about how it approximates the graph. Most lines (infinitely many) will intersect the graph at a single point. Intersecting the graph at a single point is useless as a way to describe the tangent, because that property is shared by infinitely many lines. $\endgroup$ May 24, 2019 at 18:06
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    $\begingroup$ The property you are wondering about is the wrong property to wonder about. If you care about this property but not about the actual property that characterizes tangents, then what you are doing has nothing to do with tangents, and you are purposefully looking in the wrong direction. Which is your right to do, I guess, but then don't pretend that this is about tangents, it's about you insisting on looking at the wrong thing. $\endgroup$ May 24, 2019 at 18:09

1 Answer 1


You can't prove equivalences between a precise, correct mathematical definition, and an imprecise, intuitive, incorrect attempt at a definition.

This idea that the tangent is "the line that only touches the curve once" is, to put it mildly, nonsense. As has been noted, that would mean that every line except the actual tangent is "the tangent" to the horizontal line $x=0$ at any given point. That the vertical line through the origin is the tangent to the $x$-axis at $(0,0)$, etc.

Basically, there are many serious problems with the premises behind your question. But the most important ones:

  1. It is not true that a line that only touches the graph once near the point is the tangent.

Simply note that for any function $y=f(x)$, the vertical line $x=a$ will only intersect the graph once; yet it almost never is the the tangent line to the graph. And many lines will generally only intersect the graph once: all lines between $y=x+1$ and $y=-x+1$ intersect $y=\cos(x)$ only at $(0,1)$, but none of them are the tangents to $y=\cos(x)$.

  1. It is not true that the tangent only intersects the graph once near the point.

For example, the function $$f(x) = \left\{\begin{array}{ll} x^2\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\ 0 &\text{if }x=0 \end{array}\right.$$ has $y=0$ as the tangent at $(0,0)$, but that line intersects the graph infinitely many times in any interval of the form $(-\delta,\delta)$ for $\delta\gt 0$.

Now, you may object that his is a contrived function. It doesn't matter. In fact, there's lots of function, infinitely many of them, for which 2 fails. There is no good notion of "most functions" that you can articulate that will make 2 above true for "most functions". It's just plain wrong.

Now, a better intuition is that the tangent is a straight line that "very closely approximates the graph" at and "near" the point. Can we make that precise?

Yes. Suppose $y=f(x)$ is a graph, and $(a,f(a))$ is a point on the graph. We want a line, $y=mx+b$, with two properties:

  1. $y=mx+b$ goes through the point $(a,f(a))$. For that to happen, we must have $f(a) = ma + b$.

  2. Of all lines that go through the point $(a,f(a))$, the tangent is the one that "best approximates" the graph of $y=f(x)$. That is: the relative error obtained by using the line instead of the function goes to $0$ as $x$ approaches $a$.

The "relative error" is a measure of how big the error is, relative to the size of the input. If I tell you I'm measuring a distance and I'm off by as much as five hundred meters, that's pretty good if I'm trying to figure out how far the Moon is, but it's a pretty lousy approximation if I'm trying to figure out how far the computer screen is from my face.

The absolute error in using the straight line $y=mx+b$ instead of the function $y=f(x)$ would be the distance between the value on the line and the value on the graph of the function. That is, $E(x_0) = |(mx_0+b) - f(x_0)|$.

Because we are trying to be "close to $a$", the relative error is how big this absolute error is, relative to how far we are from $a$. So the relative error at $x_0$ is: $$R(x_0) = \frac{E(x_0)}{|x-x_0|} = \left|\frac{(mx_0+b)-f(x_0)}{x-x_0}\right|.$$

Definition. The line $y=mx+b$ is tangent to the graph of $y=f(x)$ at the point $(a,f(a))$ if and only if (i) the line goes through the point $(a,f(a))$; and (ii) the relative error goes to $0$ as $x_0$ approaches $a$; that is, $\lim\limits_{x\to a}R(x) = 0$.

With this definition, it is pretty straightforward to show that the line in question must have slope equal to $f'(a) = \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$, and so must be the line $y=f'(a)(x-a) + f(a)$.

You can find the derivation and lots of discussion about this issue elsewhere on this site.

Now, you ask, can we prove that the tangent only intersects the graph at the point, at least locally, using the definition?

No. We can't because that's not true in general, as noted above. It's not true that for "most functions" (you would to specify which functions) the tangent line at $(a,f(a))$ will only intersect the graph of $y=f(x)$ at $(a,f(a))$, at least on some interval $(a-\delta,a+\delta)$; it's not true. You can't prove something that is not true, no matter how intuitively you think it ought to be true.


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