$$f:\mathbb R\rightarrow \mathbb R$$ be the function  defined by $$f(x)={{\sin x}\over {|x|+\cos x}}.$$ Then ,  $A.$ $f$ is differentiable at all $x\in \mathbb R.$  $B.$ $f$ is not differentiable at $0.$  $C.$ $f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0.$  $D.$ $f$ is not differentiable at $x={{\pi}\over 2}.$  I think option $B$ is correct as  we  know  $|x|$  is  not  differentiable  at  $x=0.$ So  that immediately  makes  options $A$  and  $C$  incorrect.   And  the  derivative can actually  be  worked  out  for  option $D$ .  So , the  correct answer  is option $B.$  Am I  right $?$   Thanks.
123870
Let $f : \Bbb R \to \Bbb R$ be the function defined by $f(x)= \frac {\sin x}{|x|+ \cos x}$. Then A.$f$ is differentiable at all $x \in \Bbb R$. B.$f$ is not differentiable at $x =0$. C.$f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0$. D.$f$ is not differentiable at $x=\frac {\pi}{2}$.  Using the standard definition of derivative, I get that $f$ is differentiable at $0$ as well as $\frac {\pi}2$. This is as follows, $\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=1=\lim_{h \to 0} \frac {f(0-h)-f(0)}{-h}$ and $\lim_{h\to 0} \frac {f(\frac {\pi}2 +h)-f(\frac {\pi}{2})}{h}=0=\lim_{h \to 0} \frac {f(\frac {\pi}2 -h)-f(\frac {\pi}2)}{-h}$ using L'Hospital rule at one stage. Hence I eliminate options B and D. Since $0$ was the only doubtful point to check about differentiability, I choose option A as the answer. I regret a little about not knowing how to check validity of option C. Can you tell me how to prove it wrong?
64075
$f:\Bbb R \to\Bbb R $ is differentiable, $f(0)=0$ and $|f'(x)|\le|f(x)|$ for all $x$ then prove $f$ is identically zero.  I tried to use mean value theorem and end up in $|f(x)|\le |x||f(c)|$ for some constant $c$ which depends on $x$. Can you help me little bit...
53982
Let $f:I\to\mathbb R$ where $I$ is an interval, $f''(x)=a \ \ \forall x\in I$.  Prove that there exsits such numbers $b,c\in \mathbb R$ such that: $f(x)=\frac a2x^2+bx+c ,\ \forall x\in I$.  Show that the above statement is not necessarily true if $f:A\to\mathbb R$ and $f''(x)=a \ \ \forall x\in A$ where $A$ is not an interval.    I'm not sure I know how to prove 1, it's probably isn't enough to just derive $f(x)$ twice  and we can't integrate... About 2. it means that the function isn't continuous but we see that it's still derivable so how can I find the relation between $b,c$ and $f$ ?
24946
Assume that $f:[a,b] \to \mathbb R$ is a function.    Prove the following statement :       If $f$ is differentiable on $[a,b]$ and $f'$ is monotonic on $[a,b]$,   then $f'$ is continuous on $[a,b]$.     Note 1 : We've learned a theorem in the class which says :       If $I$ is an open interval and $f:I\to\mathbb R$ is monotonic, then   $f$ is continuous on $I$.   So, here i can say that $f'$ is continuous on $(a,b)$. But my problem is how to prove that $f'$ is also continuous at the end points $a$,$b$.    Note 2 : There's a similar question  but please pay attention that my question is asking about when $f$ is defined on a closed interval.
104747
Let $f:\mathbb R\to \mathbb R$ be a differentiable function such that $f(0)=  0$ and $f(1)= 1$ and $|f'(x)|&lt;2 ~ \forall x \in \mathbb R$, if $a$ and   $b$ are real numbers such that the set of possible values of $\displaystyle\int_0^1  f(x)dx $ is the open interval $(a,b)$, then $b-a$ is: ?   Attempt:   $$I = \int_0^1 1.f(x) dx$$  $$\implies I = 1 - \int_0^1 xf'(x)dx $$ (Integration by parts)  $$-2 &lt; f'(x) &lt; 2$$  $$\implies -2x &lt;xf'(x)&lt; 2x$$  for $x&gt;0$  $$\implies -1 &lt; \int_0^1 xf'(x) dx &lt; 1$$  Therefore $I_{max} = 2$ and $I_{min} = 0$  $\implies b- a = 2$ but answer given is $b-a = \dfrac 3 4$.  Please let me know my mistake, and the correct way to solve it.
63795
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.
62488
Consider the real-valued function  $f(x)=x^r$ where $r$ is a real number.  (1) For what values of $x$ and $r$, is this function differentiable?  (2) How do you prove the power rule, when $f(x)$ is differentiable?  It would be nice if the proof is accessible to a 12th grade student.  EDIT: Perhaps, I have not been very clear about my question.  I am not satisfied with the "textbook proofs" I have seen.  Please established the existence (of the derivative) first before proving the power rule.  That is, please consider first question first.  This is the reason why I call the question a mathematical engineering question.  (If we can we bring high level math down to Grade 12 level, without loosing precision, then I call it good mathematical engineering.)
10069
Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $\lim_{x\rightarrow a}f'(x)$ exists. Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.    It seems like an easy example, but a little bit tricky. I'm not sure which theorems should be used in here.    ==============================================================  Using @David Mitra's advice and @Pete L. Clark's notes I tried to solve this proof. I want to know my proof is correct or not.  By MVT, for $h&gt;0$ and $c_h \in (a,a+h)$ $$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$ and $\lim_{h \rightarrow 0^+}c_h=a$.    Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$   But that's enough? I think I should show something more, but don't know what it is.
104239
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
32572
I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question:  Does there exist a continuously differentiable function, $f: [0,1] \rightarrow [0, \infty)$, such that the following hold? (i) $f(0) = 0$ (ii) $f'(x) \le c f(x)$ for all $x$ ($c$ is a fixed constant) (iii) $f \not\equiv 0$  This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance.
74232
Suppose $f(x)$ is continuous on $[a, b]$ except at a point $c$ in $(a, b)$ at which $f(x)$ has a jump discontinuity. For $x$ in $(a,b)$, set $F(x)=\int_a^xf(t) \, dt$. Show that $F(x)$ is continuous but not differentiable at $x =c$.  Hint: first explain why $f(x)$ is integrable on $[a,b]$, then consider one-sided derivatives of $F(x)$ at $x =c$.  I don't know how to step up because it seems conflict to me. $f(x)$ is not continuous because it has a jump but I have to prove it continuous, how? Can someone guide me, please.
90668
Take the function  $$ \begin{cases}f(x,y)=(x^2+y^2)/\sin(\sqrt{x^2+y^2}) &amp; \text{when } 0&lt;\lvert (x,y)\rvert&lt;\pi\\   0 &amp; \text{when } (x,y)=(0,0) \end{cases}$$  I got that this function is differentiable at $(0,0)$. I calculated the partial derivatives at $(0,0)$ of which both were $0$ and trying to show the definition of differentiability is not satisfied but I keep getting that it is. Plz help.
107615
Consider the function : $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto   \begin{cases}         0 &amp; \text{for } (x,y)=(0,0) \\    \frac{x^3}{x^2+y^2}       &amp; \text{for } (x,y) \neq (0,0)   \end{cases} $$  Show that $f$ not differentiable at $(0,0)$ but all directional derivatives exist.  I don't know how to tackle this problem. Can someone give some hints or solve the problem? Thanks :)
79431
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' &gt; 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).  Is $f$ increasing on $(a,b)$? Must $f'(c)$ be zero, or can it be negative?  Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points. And I think $f'(c)$ must be zero: If $f'(c) &lt; 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) &lt; 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) &gt; 0$, a contradiction.
55736
I came across the following problem that says:        Let $\displaystyle f \colon \mathbb R \rightarrow \mathbb R$ be differentiable function and $\displaystyle f(x)=0$ for $|x| \geq 10.$ Let  $g(x)=\sum_{k \in \mathbb Z}f(x+k).$ Then   which of the following is true?   $1.g$ is differentiable and $g'$ has infinitely many zeros   $2.g$ is continuous and $g'$ has no zeros   $3.g$ is differentiable and $g'$ has no zeros   $4.g$ is differentiable and $g'$ has only finitely many zeros.     I am not sure about how to progress with it.Can someone point me in the right direction? Thanks in advance for your time.
11457
Theorem: If $f$ is finitely differentiable at $c$, then $f$ is also continuous at $c$.  But according to the definition, for a function to be differentiable at $c$, it need not even be defined at $c$, just that it should have a finite value in the vicinity of $c$.    My question: If function is not defined at $c$, how can it be continuous at $c$?  For example, say we have a straight horizontal line function which is broken at $x=1$, the function is defined at all the points except $x=1$, it has a finite value at all the points except $x=1$, since it is having a value in the nearby region of $x=1$ we say that it is differentiable at $x=1$ but can we say that it is continuous at $x=1$?
111201
In this exercise one must show that $$f(x,y) = \begin{cases} 0 \text{     if } (x,y) = (0,0)\\ (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right) \text{     else} \end{cases}$$ is differentiable for all $(x,y) \in \mathbb{R^2}$  In the lecture, we have proven that if all partial derivatives of $f$ exist and are continuous, then the derivative of $f(x,y)$ exists everywhere.  I start by calculating the partial derivatives of $f$ at $(0,0)$: $$\partial_xf(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{x^2\sin({1 \over x^2})}{x} = 0$$ Similar for $\partial_y$: $$\partial_yf(0,0) = \lim_{y\to0} \frac{f(0,y)-f(0,0)}{x} = \lim_{y\to0} \frac{y^2\sin({1 \over y^2})}{y} = 0$$  We know that for all other values of $x$, $\partial_x f = 2x\sin\left(\frac{1}{x^2+y^2}\right)-\frac{2x\cos \left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}$  I have tried to show that $\lim _ {x,y \to (0,0)} \partial_x f = 0$. (to show f is continuous) this however is problematic, since this limit doesn't exist. This implies that the partial derivates are not continuous at all points. Which makes this approach of proving that $f$ is differentiable useless.  How do I prove that $f$ is differentiable on all of $\mathbb{R}^2$.  I am open for all kind of suggestions. Thanks for your time.