Let $f:\mathbb R \rightarrow \mathbb R$ have derivatives of all orders. Suppose that   $a&lt;b$ and that $f(a) = f(b) = f'(a) = f'(b) = 0$.       Prove that $f'''(c) = 0$ for some $c \in (a,b)$     $f(a) = f(b)$ and since f is continuous and differentiable, we can use Rolle's Theorem.  $\exists c \in (a,b)$ such that $f'(c) = 0$  So surely, differentiating twice more, $f'''(c) = 0$ for some $c \in (a,b)$?  Or am I misunderstanding something? Why would the question ask for $f'''$ ?  Thanks in advance.
121917
Let $f\colon \mathbb{R} \to \mathbb{R}$ have derivatives of all orders. Suppose that $a &lt; b$ and that $f(a)=f(b)=f'(a)=f'(b) =0$. Prove that $f'''(c) = 0$ for some $c$ in $(a,b)$.
99523
The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
122847
We are asked the following:      let $f(x)$ be a continuous function on $[0, 1]$ such that   $$\int _ {0} ^ {1} \ f(y) dy = 1$$   prove that $f(x) = 1$ for some $x ∈ [0, 1]$   The question asks to take into account Rolle's Theorem and Mean Value Theorem.  I'm aware of the definitions, but I am really unsure how to answer this question and my tutors are not really responding till after easter.    Could someone please help? Why is there a "y" involved?  Thanks you
32030
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$   and $f'(0)$ exists, then $f$ is differentiable on $\mathbb R$.   I know how to show it is continuous but no clue how to show differentiable.
81925
Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=0$ and $$\int_0^1f(x)dx=0.$$ Prove that $\exists a, b\in(0,1)$ such that $$\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$$  My approach: Let $g:[0,1]\to\mathbb{R}$ be such that $$g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt, \forall x\in[0,1].$$ By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$ Also, observe that $g(0)=0$ and $g(1)=-\int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $\exists c\in(0,1)$ such that $$g'(c)=\int_0^cf(t)dt=-\int_0^1tf(t)dt.$$ Observe that clearly three cases are possible, i.e, either$$\int_0^cf(t)dt&lt;0\text{ or }\int_0^cf(t)dt=0\text{ or }\int_0^cf(t)dt&gt;0.$$ Now let $h:[0,1]\to\mathbb{R}$ be such that $$h(x)=\int_0^xtf(t)dt, \forall x\in[0,1].$$ Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.  Observe that if $\int_0^cf(t)dt&lt;0$, then, $h(1)&gt;0$. This also implies that $\exists$ an open interval $(d,e)\in[0,c]$, such that $f(t)&lt;0, \forall t\in(d,e)$. Now select any point $c_1\in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $\exists c_2\in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)&lt;0.$$ Now $h(c_1)&lt;0$ and $h(1)&gt;0$. Thus, by IVT we can conclude that $\exists a\in(c_1,1)\subseteq(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ A similar reasoning for the case when $\int_0^cf(t)dt&gt;0,$ shows that $\exists a\in(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ Now finally if $\int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, \forall t\in[0,c]\implies h(x)=0, \forall x\in[0,c].$ Thus choosing any point $x\in(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case. Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $\exists c_1,c_2\in(0,c)$, such that $f(c_1)&gt;0$ and $f(c_2)&lt;0$. Also, let us assume WLOG that $c_2&gt;c_1$.  I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.
25084
Suppose that $f(x)$ is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. It is also   known that $|f''(x)| \le A$ for every $x \in (0,1)$. Prove that   $|f'(x)| \le A/2$ for every $x \in [0,1]$.   I'll explain what I did so far. First using Rolle's theorem, there is some point $c \in [0,1]$ so $f'(c) = 0$.  EDIT: My first preliminary solution was wrong so I tried something else. EDIT2: Another revision :\  I define a Taylor series of a second order around the point $1$: $$ f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2 $$ $$ f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2 $$ $$ |f'(1)| = \frac12 |f''(d_1)| &lt;= \frac12 A $$  Now I develop a Taylor series of a first order for $f'(x)$ around $1$: $$ f'(x) = f'(1) + f''(d_2)(x-1) $$ $$ |f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2} $$  It looks correct to me, what do you guys think?  Note: I cannot use integrals, because we have not covered them yet.
79487
Let $f$ be differentiable on $\Bbb{R}$ and let $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$. Prove there exists $a\in \Bbb{R}$ such that $f'(a)=0$.  Attempt: If $f$ is constant we are done. Otherwise, $f$ is not constant. Suppose there is no such $a$. Then, in particular, $f$ has no extremum $\implies$ $f$ has no point where is goes from increasing to decreasing or from decreasing to increasing. $f$ is continuous since it is differentiable. Hence, since $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$, $f$ must be constant. A contradiction. Therefore either $f$ is constant and $f'(x)=0$ everywhere, or $f$ has an extremum at $x=a$ and $f'(a)=0$.   I feel like my attempt is not formal enough, any correction?  I would appreciate your help.
11285
Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.
70712
Suppose that $f$ is continuous on $[0, 2\pi]$ and that $f(0) = f(2\pi)$.  Prove that there exists a $c$ in $[0, 2\pi]$ for which $f(c) = f(c+\pi)$.  Here is what i did: if $f(c) = f(c+\pi)$, then $f(c) = f(c-\pi)$.  $f(c+\pi) = f(c-\pi)$ for which $\lvert{\sin(x)}\rvert$ is the function that comes to my mind, and I'm stuck here. I must use the mean value theorem and maybe Rolle's theorem since it's a question from that section, but $\lvert{\sin(x)}\rvert$ is not differentiable. Yet the question itself doesn't mention anything about differentiability of the function. Could anyone solve this, please?
10735
I am stuck with the following problem that says :     let $f \colon [a,b] \to \mathbb{R}$ be continuous on $[a,b]$    and $f''$ exist for all $x \in (a,b)$. Let $a \lt c \lt b$.    Prove that there exists a point $\xi \in (a,b)$    such that $$f(c)=\frac{b-c}{b-a}f(a)+\left( \frac{c-a}{b - a} \right)f(b)+\frac12(c-a)(c-b)f''(\xi).$$   Can someone explain in details? Thanks in advance for your time.
100837
Problem $f(x)$ is defined over $[a,b]$ and differentiable over $(a,b)$, where $b-a\geq 4.$Prove that there exists $\xi \in (a,b)$ such that $f'(\xi)&lt;1+f^2(\xi)$. My Proof Since $b-a \geq 4$，we can obtain $$\exists x_1,x_2 \in (a,b):x_2-x_1&gt;\pi.$$ Denote$$F(x):=\arctan f(x).$$ Obviously, $F(x)$ is continuous over $[x_1,x_2]$ and differentiable over $(x_1,x_2)$. Thus, By Lagrange's Mean Value Theorem，$$\exists \xi \in (x_1,x_2) \subset (a,b):F(x_2)-F(x_1)=F'(\xi)(x_2-x_1).$$ Further, $$\frac{f'(\xi)}{1+f^2(\xi)}=F'(\xi)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\leq \frac{|F(x_2)|+|F(x_1)|}{x_2-x_1}&lt;\frac{\frac{\pi}{2}+\frac{\pi}{2}}{\pi}=1,$$ Which implies $$f'(\xi)&lt;1+f^2(\xi).$$ AM I RIGHT? HOPE TO SEE OTHER PROOFS. THX.
93845
I have been stuck for several days on this old Analysis problem (I am doing some study on my own).  I have tried several things (which I'll indicate below), but I cannot seem to figure it out.  Here is how the problem is presented: Problem: &quot;Let $f$ be a continuous real-valued function on $[a,b]$.  Suppose there exists a constant $M \geq 0 $ such that $$|f(x)| \leq M \int_a^x |f(t)| dt$$ for all $x \in [a,b]$.  Show that $f(x)=0$ for all $x \in [a,b]$.&quot; My Thoughts: I have tried using the mean value theorem iteratively, but that seems to always lead me down a dead end road.  I deduced that $f(a)=0$.  If only $f$ were assumed to be differentiable, then maybe I could play with trying to get the derivative to be $0$, but unfortunately it's only continuous.  My other thought was to (somehow) use the condition to show that $\int_a^b |f(x)| dx = 0$.  I also played around a bit with contradiction, but to no avail.  Even if one of these hair-brained thoughts is correct, I am not really sure what to do next. If you have any ideas, suggestions, or solutions, I would really appreciate it if you are willing to share them.  Thank you for your time.
6428
(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!!
81832
Please help me to solve this question:  Suppose $f:[a,b] \to \Bbb R$ satisfies:  $f''(x)+f(x)&gt;0$ and $f(x)&gt;0$ for all $x\in(a ,b)$; $f(a)=f(b)=0$.  Prove that $b-a&gt;\pi$.  Thanks in advance.
90256
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.  Define for all $x,y\in \mathbb{R}$,  $$g(x,y)=f(\sqrt{x^2+y^2})$$ Pick out the true statements :   $g$ is differentiable on $\mathbb{R}^2$ $g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$ $g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$   I do not have much idea of differentiation of multivariable calculus.  I just tried out some examples to get some idea..  I have to take some function whose derivative at $0$ has to be zero.  for simplicity I would take $f(x)=x^2$ Then I would get :     $g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$   This would suggest me that second option has more chances as $f(0)=0$ in this case.  I would now take $f(x)=x^2+1$ then :     $g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$   So, I would now eliminate second option and third option also...  So, It is now got confirmed that i have to work out to "Prove" that first option is correct.  Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.  I have to look for partial derivatives first :  $$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$  $$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$  I have used here that $f$ is differentiable on $\mathbb{R}$  As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :  $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.     As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$      As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$   So we have continuous partial derivatives for $g(x,y)$..  So, I am willing to conclude that:     $g(x,y)$ is  differentiable on $\mathbb{R}^2$   I would be thankful if some one can confirm if the answer correct.  I would be much more thankful if some one can let me know if there are any gaps in my argument.  This is just a proof verification Question..  Thank you.
38192
I am trying to prove that, if $f:\mathbb{R}\rightarrow \mathbb{R}$ and $F:\mathbb{R}^2\rightarrow\mathbb{R}$ are differentiable, $F$ satisfying $F(x,f(x))=0$ and $\partial{F}/\partial{y}\ne 0$ then $$f'(x)=-\frac{\partial{F}/\partial{x}}{\partial{F}/\partial{y}}$$  This appears in a section on the chain rule for many-variables functions, but without knowing the details of $F$ I find it very hard to prove much about it.  To the best of my ability to reason about it, I suppose I could say  $$ \frac{\partial{F}}{\partial{x}}=\frac{\partial{F}}{\partial{x}}\frac{\partial{x}}{\partial{x}} + \frac{\partial{F}}{\partial{y}}\frac{\partial{y}}{\partial{x}}$$  but for more than one reason I doubt that what I have written is correct.  Yet I'm not exactly sure what's wrong about it, nor can I guess what else I could do.  I tried finding the partial with respect to $y$ but got nothing more useful looking than this.  I also tried writing out limit definitions and saw no useful path forward through that.
122493
Given $f(x)$, which is differentiable at every point such that: $f'(x) \ge -5$ for every $ x \in R$   $f(2) = -13$, $f(9) = -48$  Prove that:$ f(3) = -18$  Now it's quite obvious that $ f(3) = -18$ since that if we know that the lowest value of the derivative is -5, the difference between $f(2)$ and $f(9)$ is $-48-(-13)=-35$  that means that that in the area $[2,9]$, $f'(x)=-5$ right? that's why I told that $f(3) = -18$, but I feel that my proof is insufficient, probably because I'm missing some theorems or such. can someone help me structure this proof into something RIGHT :)?  Thanks a lot!