We are given that a twice differentiable function, $f:\Bbb R\rightarrow[-2,2]$ satisfies the condition $$(f(0))^2 + (f'(0))^2 = 85$$  We are asked if exists a value of $x$, say $α \in (-4,4)$, for which $f(α) + f''(α) = 0$ and $f'(α) \neq0$.  (Please note we are merely asked about the existence of such value and not the value itself)  My attempt:   Suppose I take a function  $$p(x) = (f(x))^2 + (f'(x))^2$$ Taking the derivative I obtained,  $$p'(x) = 2f(x)f'(x) + 2f'(x)f''(x)$$ Which can be rewritten as  $$p'(x) = 2f'(x)[f'(x)+ f''(x)]$$  So if I can somehow prove that p'(x) = 0 at some point and f'(x) is not zero at that point, I know that there is a value of α. But not sure how to proceed any further. Can I apply LMVT somewhere? Have I missed something? Any help would be appreciated!
120673
For every twice differentiable function $f : \mathbf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statement(s) is (are) TRUE ?  (A) There exist $r, s\in \bf R$, where $r &lt; s$, such that $f$ is one-one on the open interval $(r, s)$  (B) There exists $x_0 \in (–4, 0)$ such that $|f'(x_0)| &lt; 1$   (C) $\lim\limits_{x\to\infty}f(x) = 1$   (D) There exists $\alpha\in(–4, 4)$ such that $f(\alpha) + f''(\alpha) = 0$ and $f '(\alpha)\ne0$  My attempt is as follows:-  $A)$ If function is continuous, then definitely in some part of the interval it will be increasing or decreasing hence will be one-one on the open interval $(r,s)$   $f(x)$ cannot be a constant function as in that case $f'(x)=0$ which means $f(0)=\sqrt{85}$, but it doesn't belong to co-domain of $f$  B) By mean value theorem we can say that there exists some $c\in(-4,0)$ for which $4f'(c)=f(0)-f(-4)$  Suppose if $f(0)=2$ and $f(-4)=-2$, then we can only say that there exists $c$ such that $f'(c)=1$. It will not be necessary that there exists a $c$ for which $|f'(c)|&lt;1$  C) $\lim\limits_{x\to\infty}f(x) = 1$, this is not at all necessary.  D) If function is not constant in $(-4,4)$, then we can safely say that at some point $\alpha$, $f(\alpha)\ne0$  But how to know $f(\alpha)+f''(\alpha)=0$  Actual answer is $A,B,D$ but I am able to ascertain only option $A)$. Please help me in this.
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 :)
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.
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.
73839
Let $f:\mathbb{D} \rightarrow \mathbb{D}$ be analytic such that $f(0)=0$ and $f'(0)=0$. Then which of the following are possible values of $f(1/2)=$  i. $\dfrac{1}{4}$   ii.$\dfrac{-1}{4}$   iii.$\dfrac{1}{3}$   iv.$\dfrac{-1}{3}$  By Schwarz lemma, all the options are correct. But does $f'(0)=0$ shorten the possible value list?
42084
For the function $f(x) = \sqrt{x}, \hspace{2 mm} f: [0, \infty) \rightarrow \mathbb{R}$, I wish to find all points $a$ at which $f(x)$ is differentiable, from the definition $$ f'(x) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} $$  Clearly, for our function $f(x) = \sqrt{x}$, this gives us $$ f'(x) = \lim_{x\to a}\frac{\sqrt{x} - \sqrt{a}}{x-a} $$  So the question boils down to showing for what values of $a$ this limit exists. How would this be done? Intuitively, I know that the answer is that the limit exists for all $a \in [0, \infty)$ and is given by $\frac{1}{2 \sqrt{a}}$, but I'm unsure of how to prove this from the limit definition of differentiability.
88810
Can anyone find a function $f$ with an $x$ such that $f&#39;(x)=0$, $f(x)$ is either a relative max/min and $(x,f(x))$ is an inflection point? In other words, suppose you're using the second derivative test, see that $f&#39;&#39;(x)=0$ and then note that it's actually an inflection point. Can you conclude that $(x,f(x))$ is not a relative extremum?  It's easy to find an example where $f&#39;&#39;(x)=0$ and there is an extremum--take $f(x)=x^4$, but in that case, there isn't actually an inflection point at 0.  I'm certainly open to the possibility that an inflection point can never be an extremum, though it seems like something I ought to have discovered by now, were it true.
77539
Is $f(x,y) = |xy|$ differentiable in $(0,0)$? I have no idea how to approach this problem.
11179
Let $f:R\rightarrow R$ be a function such that all its successive derivatives exist in all $R$ and also $f(x)f''(x)\leq 0$ everywhere.  If $\alpha$ and $\beta$ be two successive roots of $f(x)=0$.   Then prove that $f'''(x)=0$ for atleast one $\gamma \in(\alpha,\beta)$.  My Attempt:  I began by taking an example $f(x)=(x-1)(2-x)$ and let $x=1$ and $x=2$ be its two successive roots.The statement is trivially true.  Then I took the function $f(x)=e^{-x}(x-1)(2-x)$.The statement is true here also so on and so forth.  But what would be the exact proof I wonder. It appears to be a question of Rolle's Theorem.
71265
Suppose $f:[a,b]\to \mathbb R$ has derivative up to order $2$, and  $f'(a)=f'(b)=0$.  Prove there is  a point $c$ at $(a,b)$ such that  $$ |f''(c)|\geq 4\frac{|f(b)-f(a)|}{(b-a)^2}. $$    If it was factor 2, not 4, then I could use a Taylor expansion with Lagrange residue.
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
68197
Prove that if $f$ is differentiable on $ [a,b]$ and if $ f(a)=f(b)=0$ then for any real $k$ there is an $ x \in (a,b) $such that $$kf(x)+f'(x)=0 $$  As all the conditions of Rolle's theorem are satisfied one can say that there is at least one  $c \in (a,b) $ such that $f'(c) =0$ How should I proceed furthur ? How can I use this to get to the required equation ?
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.
98623
Let $f:(a,b)\rightarrow \mathbb{R}$.   $f$ satisfied the following property:  If $\forall x_{1},x_{0},x_{2}\in(a,b)$ and $x_{1}&lt;x_{0}&lt;x_{2};$then$\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}.$   My question : whether the function $f\in C((a,b))?$    We can get  $\displaystyle\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}. $Let $\displaystyle g(x)=\frac{f(x)-f(x_{0})}{x-x_{0}},$ if we can prove $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0});$ then we have $f^{'}_{-}(x_{0}) $ exists .In a similar way ,$f^{'}_{+}(x_{0}) $ exists.  $f^{'}_{-}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}-}f(x)=f(x_{0});$ $f^{'}_{+}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}+}f(x)=f(x_{0}).$  Obviously, $f$ is coutinuous at $x=x_{0}.$Further, $f\in C((a,b))!$     But  I failed to prove  $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0}).$Sometimes  I doubt the conculsion that  $f\in C((a,b)).$  Either make a counterexample to deny it ,or prove the conculsion is correct?
98516
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all  $x\in\mathbb{R}$, then $f\equiv 0$.  I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help?
50981
Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.   Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further.   Possibly Helpful Links: Information on similar problems can be found  and .  Source:  It is about to be closed for containing too many problems in one question. I'm posting each problem separately.
21779
Given $f:R \to R$, $f$ is differentiable on $R$ and $\lim_{x \to \infty}(f(x)-f(-x))=0$. I need to show that there is $x_0 \in R$ such that $f'(x_0)=0$  I am trying to prove it by contradiction .... so i assume there is no $x_0 \in R$ such that  $f'(x_0)=0$ this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value ....  (from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem )  now from $\lim_{x \to \infty}(f(x)-f(-x))=0$ ,we get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$  now the last step is the contradiction its very intuitive and even maybe obvious that we can't get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$ for strictly monotonic function but i cant think of a formal way to prove it .