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For example the function $f(x)=|x|$, the graphic will be something like the letter "V". The function is continuous in $x=0$. However, the slope is different for $x < 0$ than for $x>0$, just like in quadratic functions, so I don’t see how that explains it. Can’t I draw a tangent line to the graph in $x=0$ which coincides with the $x$ axis? If i am able to do that, why isn’t the derivative (in $x=0$) equal to zero as in quadratic functions?

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    $\begingroup$ Maybe related this one and this one $\endgroup$
    – mrtaurho
    Oct 8, 2018 at 14:48
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    $\begingroup$ You can draw lots of "tangent" lines at $x=0$ but there is no canonical tangent line. You are pleading for a "derivative" $0$ there, but why not e.g. $0.5$ or $-0.32$? $\endgroup$
    – drhab
    Oct 8, 2018 at 14:48
  • $\begingroup$ No, you can draw a line through the vertex point but it is not a tangent line. A tangent line requires you to put a line through 2 points on your curve then see what happens to the slope of that line as you bring one point closer to the second. $\endgroup$
    – Paul
    Oct 8, 2018 at 14:48
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    $\begingroup$ Spencer.Have a look:math.stackexchange.com/questions/991475/… $\endgroup$ Oct 8, 2018 at 14:51
  • $\begingroup$ @Paul: this cannot convince the OP, who can take two symmetrical points and confirm his theory of "slope $0$". $\endgroup$
    – user65203
    Oct 8, 2018 at 14:53

6 Answers 6

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You can think this geometrically. The derivative of a one variable function is the slope of the tangent line. The slope, which is defined as a limit, will exist and will be unique if there is only one tangent line. Now in case of $f(x)=|x|$, there is no one unique tangent at $0$. I refer to you to the following graph : enter image description here

Now ask yourself which "tangent" you want to consider?

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  • $\begingroup$ and why not say that it has infinite derivatives? $\endgroup$ Jun 5, 2020 at 21:15
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We have to define things when we discuss things.

We say $f$ is differentiable at $0$ if $\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}$ exists. If $f(x)=|x|$, we have $\lim_{x \to 0^+} \frac{|x|-|0|}{x-0}=1$ but $\lim_{x \to 0^-} \frac{|x|-|0|}{x-0}=-1$, hence the limit doesn't exists.

The notion of subgradient might be of interest to you. Subgradient of $|x|$ at $0$ is $[-1, 1]$.

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Any straight line through the origin with a slope between $-1$ and $1$ is "tangent" to the graph. So what slope are you going to take ?

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Here is the answer in the pictorial sense, not with definitions.

So what's the decisive difference between the quadratic function $g(x) = x^2$ and the absolute value function $f(x) = |x|$, such that the former is differentiable at $0$ but the absolute value is not? It's the following: Something changes for $g$ if you "zoom in" around zero, but nothing changes with $f$.

If you want to draw a tangent, it should really describe the function well in a very small region around the point. Look at the parabola: It goes up quite fast, but starts rather flatly in a neighbourhood of $0$. The closer you will look (the more you zoom in around 0), the more the curve will resemble a flat line through the origin, so it makes sense to say that the flat line is the tangent to the parabola. However, if you look more closely aroudn the origin at $f(x) = |x|$, it will always be the same V-shape and no flat line can ever appropiately approximate it. Zoom in more and more: Still, your proposed "tangent" will not come closer to the graph of f, it's not a tangent. Therefore, functions with "edges" in their graph are not differentiable there.

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I explain this with your function $f(x)=\vert x\vert$ :

A tangent line cannot be defined as a line which intersect the graph only once.

If $h \neq 0$, then the two distinct points $(0,f(0))$ and $(h,f(h))$ determine a st.line whose slope is $$\frac{f(h)-f(0)}{h}$$ The tangent line at $(0,f(0))$ is a limit of these secant lines and this case the limit does't exist(!)

For all nonzero $a$, the limit $$\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$ exist

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The reason lies within “the limit definition of a derivative”

I will jump to the explanation that sums it up since you have already stated f(x) at x=0 exists and there is continuity.

The LIMIT of that “V” shape from the left MUST equal the LIMIT of that same “V” shape from the right. When I say LIMIT, I’m referring to the limit of the slope formula itself.

The example I love to use is actually not even a mathematical “proof,” it’s something that’s way more intuitive when you really imagine it.

If you were riding a roller coaster 🎢, would you ride the one that went along a “V” track or a “U” track? Most people answer this correctly every time with indisputable reason. The v shape will surely crash because the roller coaster car never gets the chance to level out when it goes from downhill to uphill. This is a real life example of the SLOPE never equaling zero.

On the other hand, the u shape will allow the roller coaster car to eventually level out as it translations from downhill to uphill. Therefore the slope itself gets to equal zero. If you ran the scenario in reverse in your mind, the roller coaster car will always have the chance to level out on the “U” no matter if it went from downhill to uphill and vice versa.

Hope that helps!

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