[Math] How to figure out the non differentiable values of this function

calculusderivatives

Basically, I am trying to answer question B, but I am confused on what it's asking. I know that a function is differentiable at a point when there's a defined derivative at that point, but I'm confused on the non differentiable part. I have graphed the derivative by finding out the function's slope at each point, but I am not sure if that helps. Could anyone help me explain and solve part B? Thanks.

Problem

Best Answer

Intuitively, a function is not differentiable at a point if:

  • The function isn't even defined there (think $f(x)=1/x$ at $x=0$)
  • The function has a "sharp/angled point" there (think $f(x)=|x|$ at $x=0$) -- as opposed to a smooth one (compare with $g(x) = x^2$ at $x=0$). The former means you could easily draw multiple lines tangent to the function through that same point. In particular what this often means is that there is a "jump" discontinuity in the graph of the derivative.
  • The derivative "blows up" to infinity at that point (the tangent becomes vertical).

For instance, some examples:

In this example, the function $f$ is not differentiable at $x=0$.

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In this example, the function $f$ is not differentiable at $x=-2$ or $x=1$

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In this example, $f$ is not differentiable at $x=0$. This is because, not of a jump in the derivative, but $f$ not being defined there:

$$f(x) = \operatorname{sign}(x) = \begin{cases} 1 & x > 0 \\ -1 & x < 0 \end{cases}$$

Sometimes it's preferable to say that $f'(0) = \delta(0)$ in this case, where $\delta$ represents the Dirac delta function. You can probably say the same of the previous example as well, when I think about it.

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Another example: here, non-differentiability is at $x=-1$ and $x=2$. (Note these are vertical asymptotes of $f$.)

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This example is problematic because of the vertical tangent as $x=0$. Notice how the derivative blows up to infinity.

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A few cases combined here: non-existence of $f$ on $[-1,1]$, and the fact that towards each of $-1$ and $1$, the tangent becomes vertical and thus $f'$ blows up to infinity.

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With these examples and the intuition of "sharp points, vertical tangents, and discontinuities being trouble spots" in mind, you should be able to get the idea of how to solve the problem, especially if you already have a graph of your derivative.

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