Let $$f(x)=\frac{1}{16}(e^{\arctan(\frac{x}{7})} + \frac{x}{7})$$ You are given that $f$ is a one-to-one function and its inverse function $f^{-1}$ is a differentiable function on $\mathbb{R}$. Also $f(0)=\frac{1}{16}$. What is the value of $(f^{-1})'(1/16)$?
122318
Let $f\left(x\right)=x^{13}+3x^{9}+2x^{5}+x^{2}+x+2$ find $\left(f^{-1}\right)'\left(1\right)$. Can somebody show me the method? I am a little confused.
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)
101209
I think you might have to use Squeeze Theorem here? I've confused myself.      $$\lim_{x \to 0+}\frac{1}{x}-\frac{1}{\arctan(x)}$$
25585
Let $f:\mathbb R \rightarrow \mathbb R$ be a non-constant, thrice  differentiable function such that $f(1+1/n)=1$ for all $n \in \mathbb Z$.  then what is the value of $f^{\prime \prime}(1)$?   Answer is 0
91447
Problem Let $f$ be a continuous one-one function defined on an interval and suppose that $f$ is differentiable at $f^{-1}(b)$ with the derivative $f'(f^{-1}(x)) \neq 0$. Prove that $f^{-1}(x)$ is also differentiable at $b$.   I read the proof in Spivak Calculus book, but it's quite confusing to me. Then I looked up online, and I saw exact the same proof everywhere. So I wonder is there an alternative proof to this problem that I'm not aware of?   Here is the proof from Spivak book   Theorem 5 (Calculus by Spivak 4th edition, page 237-238) Proof. Let $b = f(a)$. Then     $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \displaystyle\lim_{h\to 0}   \dfrac{f^{-1}(b + h) - a}{h}$$     Now every number $b + h$ in the domain of $f^{-1}$ can be written in the form     $$b + h = f(a + k)$$     for a unique $k$. Then     $$\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h}   = \displaystyle\lim_{h\to 0}\dfrac{f^{-1}(f(a + k)) - a}{f(a + k) - b}   = \displaystyle\lim_{h\to 0} \dfrac{k}{f(a + k) - f(a)}$$     We have     $$f^{-1}(b + h) = a + k \Leftrightarrow k = f^{-1}(b + h) - f^{-1}(b)$$     Since $f$ is continuous and one-one, $f^{-1}$ is also continuous at $b$. This means that $k$ approaches $0$ as $h$ approaches $0$. Since     $$\displaystyle\lim_{k\to 0} \dfrac{f(a + k) - f(a)}{k} = f'(a) = f'(f^{-1}(b)) \neq 0$$     this implies that     $$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$$
87989
Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$. Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$. Derive a formula for $F$. Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.  If we express $S$ in polar coordinates: $$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$ then, from simple trigonometric relations, the inverse function $\Phi$ is: $$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$ and its inverse $F$ is: $$F(t)=\frac\pi{2}-2\arctan t$$ After changing coordinates back to cartesian, $F$ will look like: $$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$ which can be further simplified to: $$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$ Is this the formula of $F$? Is it correct? And I know that to show that $\phi$ : $\mathbb R$ →  $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me show that $\phi$ is bijective? and how can I derive a formula for its inverse?
67960
I proved as follows:  Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$ Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x &gt; \tan x - x,$ since $\\tan x &gt; x,$ $f''(x) &gt; 0$ so $f'(x)$ is increasing. Since $f '(0) = 0,$ $f '(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$ So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$  But my proof seems complex. Is there any simpler way to prove this?
47323
Using $$\frac{f(x+h)-f(x)}{h}$$ how would I find the derivative of $f(x)= \sqrt{x}+1$   The answer is $$\frac{1}{2\sqrt{x}}$$ but I keep getting $$\frac{1}{2 + 2\sqrt{x}}$$
63149
Let $f$ be a twice differentiable function on the open interval $(-1,1) $such that $f(0)=1$. Suppose $f$ also satisfies $f(x) \ge 0, f'(x) \le 0 $and $f''(x) \le f(x)$, for all $ x\ge 0$. Show that $f'(0) \ge -\sqrt2.$  My attempt $ \rightarrow$  From Taylor's we see that $0 \leq f(x)=f(0)+f'(0)x+\frac{f''(\zeta)x^2}{2} $ for some $\zeta \in (0,x) .$  $f'(x)\leq 0$ and $  f'(x) \le 0 \implies$  $  f''(\zeta) \leq f(\zeta)\leq 1$  Thus $1+f'(0)x+\frac{x^2}{2} \geq 0 $.  I'm stuck upto this,I can see that if the discriminant of the quadratic is less than $ 0$ we're done but how do I conclude that the discriminant would be $0$ ?
109538
Let $f: \mathbb{R}\to \mathbb{R}$ be a twice differntiable function and $f(\frac{1}{n})=0$ all $n\in \mathbb{Z^{+}}$? then which of the statement (s) are correct from below.  (a)$f(0)=0$  (b)$f^{'}(0)=0$  (c)$f^{''}(0)=0$  (d)$f$ is a non zero polynomial  I tried of using Tyler series but I wasn't able to do this any help would be appreciated thank
99220
Find $f$ if $$ f(x)+f\left(\frac{1}{1-x}\right)=x $$ I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$ I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.
101507
I am trying to find a function such that   $\dfrac{d}{dx}f(x)=f(x-1)$   Is there such function other than $0$ ?
52787
How do I show this?  $$\cos^2(\arctan(x))=\frac{1}{1+x^2}$$  I have absolutely no idea. Thank you.
53559
I'm having problem with showing that: $$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$  I would need some help in the right direction
72025
Let $f(x)$ be a function which is differentiable on $[0,1]$ with $f(0)=0$ and $f(1)=1$. Show that for every $n\in \Bbb N$ there exists numbers $x_1,x_2,\ldots,x_n\in [0,1]$ such as $$  \sum_{k = 1}^n \frac{1}{f' (x_k)} = n $$ I think the mean value theorem should be applied. So there exists $x_1$ in $[0,1]$  such that $f ' (x_1) = \frac{f(1) - f(0)}{1-0} =1$  and there exists $x_2$ in $[0,x_1]$ such that $f ' (x_2) = \frac{f(x1) - f(0)}{x1-0} = \frac{f(x1)}{x1}$, so on and so forth for $x_3 ,x_4, \ldots,x_n$ and we have the sum $$1+\frac{x_1}{f(x1)} + \frac{x_2}{f(x2)} +\cdots+\frac{x_{n-1}}{f(x_{n-1})}$$  and from here I have no idea what to do . I was wondering if anyone could be so kind to help ?
29487
Problem : Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable?   For the sake of generality, let's assume that it is unknown to us that $|x|$ is not differentiable at $x = 0$  Attempted Solution :  Using the definition of differentiability, a function is differentiable over an interval $I$ $\text{iff}$  $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \ \ \forall x\in I$$  Now implicitly this definition of differentiability requires $$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}$$  Therefore $f$ will be differentiable only when  $$\lim_{h\to 0^+} \frac{(x + h + |x+h|)-(x + |x|)}{h} = \lim_{h\to 0^-} \frac{(x - h + |x-h|)-(x + |x|)}{h}$$  But it is unclear what to do next as $x$ is just arbitrary $$ \begin{equation} \begin{aligned} |x+h| &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq -h\\ -x-h &amp; \text{if} &amp; x &lt; -h \end{cases} \\ &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq 0\\ -x-h &amp; \text{if} &amp; x &lt; 0 \end{cases} \end{aligned} \end{equation}$$    A Wrong Solution :  We could take the derivative of $f$, (and evaluate the domain of the derivative $f'(x)$ to values for which it is defined), using the rules of differentiation, but we would get a wrong answer.  $$f'(x) = \frac{|x| + x}{|x|}$$ $$\implies f'(x) = \begin{cases} 2 &amp; \text{if} &amp; x \geq 0 \\ 0 &amp; \text{if} &amp; x &lt; 0 \end{cases}$$  This implies the derivative $f'(0)$ exists, when it does not as per the definition of differentiability. Why is that so?    Questions:   How can a solution be found using the definition of differentiablity? Why does taking the derivative of $f$, $f'(x)$ and evaluating it's domain, not give the correct values of $x$ for which $f$ is differentiable?
67053
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.   My  attempt: $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$ $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$ $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$ $\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$ $\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$ As denominator is tending to zero,so numerato will also tend to zero. $1+a-b=0............................(1)$ Applying L Hospital rule, $\lim_{x\to 0}\frac{-a\sin x}{2x}=1$ So $a=-2$ and $b=-1$ But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$. I do not understand where have i gone wrong?