Is it in general true that for $f(x)$ continuous also $\frac{f(x)}{f'(x)}$ is continuous?
If not, are there certain circumstances under which it is?
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You need to say $f$ is differentiable to start with, then the continuity of $f$ is automatic. – zhw. Nov 09 '17 at 21:45
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You can have $f(x)$ continuous and $f'(x)$ not. A simple example is $f(x)=|x|$ where $f'(x)$ is discontinuous and undefined at $x=0$. Any curve with "corners" will suffer this fate. Any time $f'(x)$ is continuous and nonzero, $\frac {f(x)}{f'(x)}$ will be continuous. The fact that $f'(x)$ exists is sufficient to guarantee that $f(x)$ is continuous.
Ross Millikan
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1So to show that $\frac{f(x)}{f'(x)}$ is continuous, I have to show that that $f'(x)$ is and that is enough, right? – Vazrael Nov 09 '17 at 21:33
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Yes, if $f$ and $g$ are both continuous so is $f(x)/g(x)$. – Henrik supports the community Nov 09 '17 at 21:35
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1Wrong. The derivative of $x\mapsto |x|$ is defined on $\mathbb R\setminus{0}$ and it is continuous. – CiaPan Nov 09 '17 at 21:50
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2This answer is incorrect. $f(x)=7$ is a continuous function with a continuous derivative, but the ratio isn’t defined for any $x$! You need for $f’(x)$ to be continuous and non-zero. – Stella Biderman Nov 09 '17 at 21:52
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@StellaBiderman: you are correct. I have updated the answer. Thanks – Ross Millikan Nov 10 '17 at 00:06
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Just because $f$ is continuous, it doesn't need to be differentiable, i.e. $f'$ might not be defined. In that case $\frac{f(x)}{f'(x)}$ is not defined either, much less continuous.
$f$ can also be differentiable with a non-continuous derivative.
Other questions on this site contain plenty of example of functions that are weird in any way imaginable, just search for it.
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If a derivative (of a real function) exists on some open interval, doesn't it have to be continuous on that interval? – CiaPan Nov 09 '17 at 21:48
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@CiaPan If "it" refers to the function, then yes. If "it" refers to the derivative of the aformentioned function, then no. – Nov 09 '17 at 22:25
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@G.Sassatelli I'm afraid both my imagination and my English are not good enough for that. Luckily I managed to find an example: $f(x)=x^2\cdot\sin(1/x); f(0)=0$ is differentiable with $f^\prime(x)$ oscillating between $-1$ and $+1$ in each $(-\epsilon,+\epsilon)$ interval whilst $f^\prime(0)=0$, hence $f^\prime$ exists in $\mathbb R$ but is discontinuous at $x=0$. Same for $g(x)=x^2\cdot\sin(1/x^2); g(0)=0$ and even more: $g^\prime(0)=0$ but $g^\prime(x)$ oscillates between $-\infty$ and $+\infty$ in each neighborhood of $x=0$... – CiaPan Nov 10 '17 at 08:13
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The answer is NO.
As others already said, $f$ being continuous does not imply $f$ is differentiable, hence $\frac {f(x)}{f'(x)}$ needs not be even defined.
For example the Weierstrass function is continuous, but not differentiable. In this case $f/f'$ is not even defined, let alone continuous...!.
Additionally, even if $f'(x)$ exists it may be zero, in which case $\frac {f(x)}{f'(x)}$ is undefined, too.
Let $f(x)=1$. It is continuous on $\mathbb R$. Then $f^\prime(x)=0$ is continuous, too.
However, $$\frac{f(x)}{f^\prime(x)}=\frac 10$$ does not exist, either
Finally, as Henrik points out in the answer, even if $f'(x)$ exists and is non-zero, it may be discontinuous, hence $\frac {f(x)}{f'(x)}$ needs not be continuous.
CiaPan
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