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I've searched online and found almost nothing. What in the mathematical definition of a derivative makes it so that the derivative of the following is undefined at 0.

\begin{equation*} f(x) =\begin{cases} 3x & \text{if } x < 0 \\ 3x+2 & \text{if } x \ge 0 \end{cases} \end{equation*}

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Who says it's undefined at 0? (This is a serious question, not rhetorical.) –  MJD Jun 27 at 21:17
    
@MJD My textbook. –  Mike G Jun 27 at 21:18
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The left limit $$\lim_{h\rightarrow0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h\rightarrow0^{-}}\frac{3h‌​-2}{h}=+\infty$$ –  Dario Jun 27 at 21:23
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Although the derivative in 0 does not exist, the left derivative and right derivative in 0 do (and are equal). In fact we are dealing with a simple example of a function of bounded variation. Such functions have many special properties, see Rudin's book on analysis. –  Urgje Jun 27 at 22:21
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4 Answers 4

up vote 2 down vote accepted

Here's my intuitive argument for why your book says what it does:

I personally consider the derivative as the best linear approximation of a function at some given point. For this function, at $x=0$ there isn't any good sort of linear approximation so we cannot have such a derivative.

As for a more proofy argument consider the limit $$ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $$ which I'm sure you recognize as the derivative. Now note that we must have this limit exist no matter how we approach it, i.e. if $h \in \mathbb{R}^-$ or $h \in \mathbb{R}^+$ or some combination of those two statements. Let's first consider when $h$ is negative and try to evaluate what $f'(0)$ is: $$ f'(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{3h - 2}{h} = + \infty $$ so you can see that this does not exist.

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A function is continuous at a point if it is differentiable at it. But here you have a piecewise function, that is discontinuous at $0.$

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One thing that you should always note is that if $f(x)$ is differentiable at $x=a$ then $f(x)$ is continuous at $x=a$. Thus this means if it is not continous at $x=0$ it can't be differentiable at that point either.

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Recall the following definition of a derivative: $$f'(x) = \lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\text{.}$$ This limit exists if and only if the two-sided limits exist and $$\lim\limits_{h \to 0^{-}}\dfrac{f(x+h)-f(x)}{h} = \lim\limits_{h \to 0^{+}}\dfrac{f(x+h)-f(x)}{h}\text{.}$$

Edit: My attempt of computing the limits: $$\begin{align} &\lim\limits_{h \to 0^{-}}\dfrac{3(0+h)- 3(0) - 2}{h} = \lim\limits_{h \to 0^{-}}\dfrac{3h-2}{h} = +\infty\\ &\lim\limits_{h \to 0^{+}}\dfrac{3(0+h)+2-3(0)-2}{h} = \lim\limits_{h \to 0^{+}}3 = 3\text{.} \end{align}$$

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But don't those two limits 0- and 0+ evaluate to the same thing since the +2 cancels out? –  Mike G Jun 27 at 21:20
    
Hmm. Good point. Maybe my alternative argument is correct? –  Clarinetist Jun 27 at 21:21
    
In your case $ \lim\limits_{h \to 0^{+}}\dfrac{f(x+h)-f(x)}{h}$ = $ \lim\limits_{h \to 0^{+}}\dfrac{2-0}{h}$ which is infinite . –  Tom Collinge Jun 27 at 21:23
    
I think I figured it out. What I just said is wrong; it doesn't cancel out. From the positive side, the limit uses the function f(x) = 3x+2 and f(x+h) = 3(x+h) + 2, but from the negative side it uses f(x) = 3x and f(x+h) = 3(x+h) +2. That +2 doesn't cancel. –  Mike G Jun 27 at 21:24
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@TomCollinge how do you figure that? The right handed limit is $$ \lim_{h \to 0^+} \frac{3x + 3h + 2 - 3x - 2}{h} = 3 $$ I believe the left hand sided limit is where problems arise. –  DanZimm Jun 27 at 21:26
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