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Perhaps my definition of 'tangent' is the problem but in school the tangent is always defined as a line that intersects with a curve at only one point. According to this definition the equation $y = x^2$ has two tangents at the point x = 0:

1) $x = 0$ (the $y$-axis)

2) $y = 0$ (the $x$-axis)

Is this true or am I missing something?

maydc
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  • Your definition is defective because it has exceptions such as you provided. – Somos Mar 15 '19 at 21:11
  • Related to MSE queston 2850391 "Another way to do differentiation? When is it applicable?" – Somos Mar 15 '19 at 21:16
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    It turns out that defining a tangent is not so straight forward. See this for an example of how elusive it can be. – John Douma Mar 15 '19 at 21:16
  • Another example of where the definition you've been used to breaks down is the tangent to $\cos(x)$ at $x = 0$ - it's the line $y = 1$ which meets the graph of $\cos(x)$ at infinitely many points. The definition you've been given works for some special kinds of curves like circles and ellipses, but not in general. – Rob Arthan Mar 15 '19 at 21:16
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    One should distinguish between "tangent" (from Latin tangens, "touching") and "secant" (from Latin secans, "cutting") – MPW Mar 15 '19 at 21:17
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    If that's your definition of tangent, then any vertical line satisfies it so it's not true that the equation has only two tangents. I'll edit to include MathJax formatting. In the future, please do so yourself. It's considered a courtesy to other contributors. – Robert Shore Mar 15 '19 at 21:21
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    A better (but imprecise) definition of a tangent line to a curve at a point is "the line that the curve looks like when you zoom-in really-really-really-...-really close to that point". – Blue Mar 15 '19 at 21:26
  • If you think of a parabola as a degenerate ellipse, with one focus carried off to infinity, then the line intersects the ellipse at infinity... – Jair Taylor Mar 15 '19 at 21:28
  • @RobertShore oh...right. I guess this is true for literally every function. Will make sure to format correctly in the future. – maydc Mar 17 '19 at 06:09

2 Answers2

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Tangent lines look like this:

enter image description here

The red line is tangent to the black curve at the red point. It is not tangent to the curve at the unmarked point where the red line stabs through the curve.

As others have pointed out, your definition of tangent is missing something. Saying the line $L$ is tangent to the parabola $y=x^2$ at the point $a$ means two things:

  1. The line meets the point $a$.
  2. There is a small horizontal region around $a$ such that, inside that region, $L$ is contained entirely above or below the parabola.

In this case one such horizontal region looks like this:

enter image description here

You cannot do this when the curve is $y=x^2$ and $L$ is the x-axis, since every horizontal region will contain all of $L$ and thus contain parts above and parts below the curve.

If you ask your teacher, I'm sure they will tell you the definition of tangent includes this extra condition.

Daron
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    This attempt to provide intuition can be misleading. Consider $y=x^3$. In any neighborhood of $(0, 0)$, part of the curve is above the tangent line $y=0$ and part is below it. – Robert Shore Mar 15 '19 at 21:42
  • Oooh! You're right! – Daron Mar 15 '19 at 21:44
  • I will leave the (incorrect) answer up to encourage others to give a correct one! – Daron Mar 15 '19 at 21:48
  • Well, this user does mention the fact that the tangent line should be ether entirely above or below the curve near the point of tangency. That condition fails for $x=0$ because the line is both below and above the curve $y=x^2$ near the point of tangency. Am I right? – Michael Rybkin Mar 15 '19 at 22:04
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I think that fundamentally, the correct definition of tangent should say something about how good the function can be approximated by a linear function in a neighborhood around the point.

More precisely, we may say that the line defined by $y = Ax + B$ is a tangent to the function $f$ at a point $x_0$ if the following two conditions hold:

(1) $Ax_0 + B = f(x_0)$

(2) $f(x_0 + h) - f(x_0) - Ah = o(h)$

Condition (1) says that the tangent line should cross the graph of $f$ at the point $(x_0, f(x_0))$.

Condition (2) says that the error between $f$ and the linear approximation given by $Ax + B$ becomes arbitrarily small relative to the distance $h$ from $x_0$ (in both positive and negative directions). The $o(h)$ term is Little-O notation.

Note that $A$ is uniquely defined by condition (2) (if any such $A$ exists), and $B$ is then uniquely defined by condition (1). So there can only ever be one tangent line to $f$ at any given point. It is also possible that no tangent line exists, such as for functions like $f(x) = |x|$ at $x_0 = 0$, or other non-differentiable functions.

We see that for the parabola, the line $y = 0$ satisfies both conditions at $x_0 = 0$, but the line $x = 0$ does not satisfy condition (2). Intuitively, at any point other than $x_0$, the line $x = 0$ will be a horrible approximation for the parabola $y = x^2$, since the line does not even pass through any points with nonzero abscissa. Actually, it does not really fit the condition (1) either because the way I've formulated things, the line $x = 0$ is not of the form $y = Ax + B$ (you would need $A = \infty$). This can be generalized a bit more by considering lines of the form $Ax + By = C$ instead, which would allow representing vertical lines as well.

Note that condition (2) can be reformulated as saying that $$\lim\limits_{h\to0}\frac{f(x_0 + h) - f(x_0)}{h} = A$$ which is the usual definition of differentiability for $f$ at $x_0$ (with derivative $f'(x_0) = A$).

The way that I formulated (2) above allows an easier (and, in my opinion, more intuitive) generalization of tangent planes for multivariable functions. In that more general setting, $f$ is a function in $\mathbb{R}^m\to\mathbb{R}^n$, $\mathbf{x}_0$ and $\mathbf{h}$ are now vectors in $\mathbb{R}^m$, and $A$ is a linear transformation in $\mathbb{R}^m\to\mathbb{R}^n$ ($n \times m$ matrix). The error is compared relative to $|\mathbf{h}|$.

Tob Ernack
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