Is there a function which is continuous and differentiable, but is not smooth function?    By smooth I mean having continuous derivative. For example, the derivative of $f(x)=x|x|/2$ is $f'(x)=|x|$ which is continuous. So I consider this function  smooth.
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If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?  My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable.
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Let $f: \mathbb{R} \to \mathbb{R}$ be given. Assume that the square and cube of $f$ are smooth. Is $f$ smooth? That is if $f \cdot f \in C^{\infty}$ and $f \cdot f \cdot f \in C^{\infty}$, does it follow that: $f \in C^{\infty}$   I got this from another question on SE.   So, $f(x)^2$ is infinitely differentiable, and so is: $f(x)^3$   Also, realize:  $$f(x) = \frac{f(x)^3}{f(x)^2}$$  But what can I do?
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My teacher showed us this function and told us it was continuous at all non-$\mathbb{Q}$ points:  $$ f(x) = \begin{cases}   x &amp; \text{ if } x\in\mathbb{Q} \\\\   0 &amp; \text{ if }x\notin\mathbb{Q} \end{cases} $$  However, Wolfram MathWorld says , which is very similar, is discontinuous at all points. Why is one continuous and the other not?
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Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
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Is there a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) = 0$ when $x &lt; 0$ and $f'(x) = 1$ when $x \geq 0$?   Attempt: I was trying to think of an example. First thing that came to mind was $$ f(x) =  \begin{cases} x &amp; x \geq 0 \\ 0 &amp; x &lt; 0 \end{cases}. $$ I wanted to know if it is differentiable at $x = 0$. Since $$ \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = 0 = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x} $$ the derivative at $x = 0$ exists and equals $0$. But according to the problem, the derivative should be $1$, so I guess this is not a correct example. Does anyone have an example, or is such a function not possible at all?
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(a) Let a function $g:I\to\mathbb{R}$ be continuous at $x_0\in I$. Prove that the function $f$ given by $f(x)=(x-x_0)g(x)$ is differentiable at $x_0$.  (b) Let a function $f:I\to\mathbb{R}$ be differentiable at $x_0\in I$ and $f(x_0)=0$. Prove that there exists a function $g:I\to \mathbb{R}$ that is continuous at $x_0 \in I$ such that $f(x)=(x-x_0)g(x)$.    Thought:  (a)  By $Fermat's$, since $g$ is continous, if $g'(x_0)=0$ then $g(x)$ is differentiable?  but how does it connect back with $f(x)$?  (b)  use the definition of every differentiable function is continuous?  Thus $f(x)$ is continuous and also $(x-x_0)$ so $g(x)$ too?  Thank you!!
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If a function is differentiable and monotone on the interval $(a, b)$, then its derivative is also monotone on $(a, b)$.   How do you prove this statement is wrong?   Can you please provide an example?
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I need to  Let $f : (a, b) → R$ be differentiable on $(a, b)$. Prove that $f$ is continuous on $(a, b)$ by showing that:  $∀x_0 ∈ (a, b)$ $∀ε &gt; 0$ $∃δ &gt; 0$ s.t.  $(x ∈ (a, b), |x − x_0| &lt; δ) ⇒ |f(x) − f(x_0)| &lt; ε$  My understanding so far:   When we say a function is differentiable at $x_0$, we mean that the limit: $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  My attempted proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  My issue:  My issue may be a non issue, however for the 6 marks this question should be worth my proof seems shallow. Could someone please poke some holes in my proof, or add to it?   Thank you
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How would I go about showing that if $f$ is uniformly differentiable then $f'$ is continuous.  my attempt: A differentiable function $f:[a,b]\to \Bbb R$ is said to be uniformly differentiable on $[a,b]$ if $\forall \epsilon&gt;0 \exists \delta&gt;0:\forall x,y\in[a,b]$ we have $$0&lt;|x-y|&lt;\delta\implies |\frac{f(x)-f(y)}{x-y}-f'(y)|&lt;\epsilon$$  I don't know where to go from here, any help would be highly appreciated.
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I was reading about one one functions and found out that they cannot have maxima or minima except at endpoints of domain. So their derivative , if it exists,  must not change it sign , i.e. , the function should be either strictly increasing or strictly decreasing. From this I've a feeling that all continuous one one functions must be differentiable . Is this true?
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I need to find a pair of functions $f$, $g$ such that    $f$ is not differentiable at  $x = 0$ $g$ is not differentiable at $f(0)$ $g \circ f$ is differentiable at $x = 0$   I've tried a lot of functions but just can't seem to find ones that work. Any help would be appreciated, thank you.
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If $|f(x)|$ is continuous at $a$, is $f(x)$ continuous at $a$?  I tried doing it using composite functions. If $g(x)= |x|$, then $g\circ f(x)= |f(x)|$. Since $g(x)$ and $g\circ f(x)$ are continuous, $f(x)$ is continuous.  I don't know if this is correct. Please help.
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Is the function $f(x) = \lim_{k \to +\infty} \tan{kx}$ a right example?  But I do not know whether this is a function at first. Is this a function?  Could you give a correct real-valued function that is continuous at precisely one point?  Thanks.
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Let $f(x)$ be a real-valued differentiable function on an interval $I$ of $\mathbb R$, then is $f'(x)$ necessary to be continuous? I don't think so and I'm trying to construct a counterexample. $\sin\frac{1}{x}$ may be the key element but $x\sin\frac{1}{x}$ failed me.
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So. I have to show that this functiun is differentiable outside of $(0,0)$ and also to calculate  its differentiation. $$f(x,y)=xy\frac{x^2-y^2}{x^2+y^2}$$ when $(x,y)$ is not equal to $(0,0)$ and $$f(x,y)=0$$ when $(x,y)$ is equal to $(0,0)$. Also we have to show that the function is differentiable on $(0,0)$ too. $$$$ Firstly I wanted to see if the functiun is continuous e.g: $$\lim_{(x,y)\to (0,0)}{xy\frac{x^2-y^2}{x^2+y^2}}=0$$ so it is countinous.$$$$ Than I wanted to see if it is differentiable but I don't know how to continue, all I got is: $$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-f(0,0)}{(x,y)-(0,0)}=\lim_{(x,y)\to (0,0)}\frac{xy\frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}$$ Can someone put me on the right track? And explain me what am I doing wrong? I don't know how to continue it..
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Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.  I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?
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I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$    Differentiability Definition  When we say a function is differentiable at $x_0$, we mean that the limit:  $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  Continuity Definition  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.  Proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.  My Question  The proof seems to execute the following steps:   Assume the function is continuous at $x_0$ Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.   What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.  Hopefully my confusion is clear. Any help will be greatly appreciated.