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I have to find a differentiable function $f: \mathbb{R} \to \mathbb{R}$ with $f'(x)=0$ if $x < 0$ and $f'(x)=1$ if $x≥0$.

I think that such a function doesn't exist because the left and right limit for $x \to 0$ are different. Can I proof it like this or is there something that I'm missing?

Mee98
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  • Good question. The answer lies in the mean value theorem - use it to prove that $f(x)$ is constant for $x \le 0$, and then, as you suggested, show that the left- and right-handed derivatives of $f$ at $x = 0$ cannot agree. – Chris Jun 14 '17 at 19:12
  • $f(x) - x$ is constant for $x \geq 0$. – user49640 Jun 14 '17 at 19:20

2 Answers2

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The function is such that $$ \lim_{h \to 0 ^-}\frac{f(0+h)-f(0)}{h}=0 $$ and$$ \lim_{h \to 0 ^+}\frac{f(0+h)-f(0)}{h}=1 $$ so $$ \lim_{h \to 0 }\frac{f(0+h)-f(0)}{h}=f'(0) $$ does not exists and the function is not differentiable at $x=0$.

Emilio Novati
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$f'=0$ at $(-\infty,0) $ means

$f (x)=C $ for $x <0$.

$f'=1$ at $(0,+\infty) $ gives

$f (x)=x+D $ for $x>0$.

to be continuous at $x=0$, we need

$$C=D $$ to be differentiable at $x=0$,

$$\lim_{0^-}\frac{C-C}{x}=\lim_{0^+}\frac {x+C-C}{x} $$

or $$0=1$$ which is not possible.

hamam_Abdallah
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