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Let $f(x)=\begin{cases}\frac{1}{2}x^2 ~~~~~&\text{if } |x|\leq c \\ c|x| - \frac{1}{2}c^2 ~~~~~&\text{if } |x|> c \end{cases}$, where $c>0$ is just a constant value in $\mathbb{R}$

Then I find out $f'(x) = \begin{cases}x ~~~~~&\text{if } |x| < c \\ \frac{cx}{|x|} ~~~~~&\text{if } |x| > c \end{cases}$.

Since, $c>0$ is a constant in $\mathbb{R}$.
Then, $\lim_{h\rightarrow c^{-}}\frac{f(h)-f(c)}{h}=\lim_{h\rightarrow c^{-}}\frac{\frac{1}{2}h^2-\frac{1}{2}c^2}{h} = \frac{0}{c} = 0$
Also, $\lim_{h\rightarrow c^{+}}\frac{f(h)-f(c)}{h}=\lim_{h\rightarrow c^{+}}\frac{c|h|-\frac{1}{2}c^2-\frac{1}{2}c^2}{h} = \frac{0}{c} = 0$

So, in this case is $f'(x)=\begin{cases}x ~~~~~&\text{if } |x| < c \\ \frac{cx}{|x|} ~~~~~&\text{if } |x| > c \\ 0 ~~~~~&\text{if } |x| = c \end{cases}$ the derivative of $f(x)$ above ?

xxxxxx
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1 Answers1

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It does not seem quite right to me. The denominator should be $h-c$ in each calculation.

For $x = c > 0$, the left derivative is, \begin{align} \lim_{h \to c^-} \frac{f(h)-f(c)}{h-c} = c \end{align} and the right derivative, \begin{align} \lim_{h \to c^+} \frac{f(h)-f(c)}{h-c} = c \end{align} so that the derivative at $c$ exists but is $c$, not $0$. Similar arguments apply when $x = -c$.

WA Don
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  • So, is that mean that the derivative of $f(x)$ shoud be $f'(x)=\begin{cases}x ~~~~~&\text{if } |x| \leq c \ \frac{cx}{|x|} ~~~~~&\text{if } |x| > c \end{cases}$ ? – xxxxxx Oct 12 '21 at 01:47
  • Yes; moreover you could add the 'or equal' part to either inequality since both give the same value when $x = \pm c$. – WA Don Oct 12 '21 at 07:01