# Tag Info

## New answers tagged functions

0

$$f1=f2\iff 45.7\cdot x-600=-0.426\cdot x+240 \iff 45.7\cdot x+0.426\cdot x=240+600$$ $$\iff x=\frac{840}{46.126}\approx18.211.$$ So your functions intersect at the point with x-coordinate $18.211$. The function your plot2 shows is thus $$f(x) = \begin{cases}f_1(x)\ \mbox{ if }\ x\leq18.211\\ f_2(x)\ \mbox{ if }\ x>18.211 \end{cases}$$ How to plot this ...

0

Let $f_1=ax+b$ and $f_2=cx+d$ with $a \neq c$, then what you call Plot2 is given by the graph of the function $f$ defined as follows: $f(x)=f_1(x)$ for $x \leq (d-b)/(a-c)$ and $f(x)=f_2(x)$ else.

1

In fact, without any hypothesis of continuity, one can make other solutions. To see one, let $A=\mathbb{Q}(\sqrt{2}) =\{a+b\sqrt{2}, a,b\in \mathbb{Q}\}$. We have for $x\in \mathbb{R}$ that $x\in A$ is equivalent to $\displaystyle \frac{x}{\sqrt{2}}\in A$. Hence if we define the function $g$ by $g(x)=2^{x^2}$ for $x\in A$ (hence as $1\in A$, we have ...

0

Write $$f(x)=f^2\left(\frac{x}{ \sqrt 2}\right)=f\left(\frac{x}{ \sqrt 2}\right)f\left(\frac{x}{ \sqrt 2}\right)$$ in table \begin{align} f\left(\frac{1}{ \sqrt 2}\right)&=&\sqrt{2}\\ % f(1)&=f\left(\frac{1}{ \sqrt 2}\right)f\left(\frac{1}{ \sqrt 2}\right)&=2\\ % f(\sqrt{2})&=f\left(\frac{\sqrt{2}}{ \sqrt ... 0 We begin by noting that the sets P_i = \{ x| \frac{i(i-1)}{2} < x \le \frac{i(i+1)}{2} \} are disjoint. Now note that \frac{(x+y)(x+y+1)}{2}+y is in the set P_{x+y+1}. Similarly, we note that \frac{(x'+y')(x'+y'+1)}{2}+y' is in the set P_{x'+y'+1}. Therefore, we must have x+y=x'+y'. It follows that x=x', y=y'. 1 f(x)=\frac{x^7}{a^8-x^8}=\frac{x^7}{a^8}\frac{1}{1-(\frac{x}{a})^8}=\frac{x^7}{a^8}\sum_{i=0}^\infty(\frac{x}{a})^{8i} with convergence interval as |\frac{x}{a}|<1 0 u^{\frac12 + \frac14 + \frac 18 + \cdots} \rightarrow u for all u and 2^n \rightarrow \infty so the bound is infinite if u > 0 and 1+0 if u = 0. 4 One systematic way to do this is to find such a function satisfying:f\left(x+\frac{2}5\right)=f(x)+1$$Once you've defined it over the interval [0,\frac{2}5] the rest of the values can be found from this relation. The only constraints on f in that interval will be that f(\frac{1}5)=f(1)-2=f(0)-2 and that f(\frac{2}5)=f(0)=1. Any continuous f ... 3 If two functions have the same output and the same domain, they are the same function. You know that g(x)=kx+n, and that g^{-1}(x) = g(x)=kx+n. Now, using the properties of inverse functions, you also know that for each x, you have x = \mathrm{id}(x) = (g\circ g^{-1})(x) = (g\circ g)(x)=g(g(x)), and this equation should give you a lot of ... 3 So we have x^2 = 2^x. Taking the square root of both sides and assume the solution is negative gives x=-\sqrt{2}^x. We can then establish a recursive sequence, x_n = -\sqrt2^{x_n-1}. Assuming that this converges gives us the answer, x=-\sqrt2 ^{-\sqrt2 ^ {-\sqrt2 ^\cdots}}. After five iterations, we get$$x\approx-0.76961847524.$$Substituting the ... 7 Suppose , gcd(a,b)=1 , x=\frac{a}{b} and x^2=2^x We have$$a^2=b^2\times 2^{a/b}$$implying$$a^{2b}=b^{2b}\times 2^a$$This is impossible, if gcd(a,b)=1 and b>1. The negative solution is obviously not an integer. If x is irrational algebraic, then 2^x is transcendental, but x^2 is not. So, x , the negative solution, must be a ... 1 Can we say that if |supp (f)| = |Domain (f)|, f(n) is finite? This question is very confused. First of all, if f is a function from \mathbb N to \mathbb N, then for every n, the number f(n) is an element of \mathbb N, and therefore finite. Second of all, as far as I see in your example, the domain of f is always \mathbb N, meaning that ... 0 Let a_k ,k=1,\cdots,10, the given zeros of f, suppose that 0<|a_1|\leq ...\leq |a_{10}|=L\leq 100. Define g holomorphic by f(z)=g(z)\prod(z-a_k). We have g(0)a_1\cdots a_{10}=1. Now for |z|=300, we get with M<1024$$|g(z)|\prod (300-|a_k|)\leq |f(z)|\leq M$$Hence \displaystyle |g(z)|\leq \frac{M}{\prod (300-|a_k|)} on |z|=300. ... 1 Hint:$$\tan(A+B)=\frac{\tan A + \tan B}{1-\tan A\tan B}$$Let A=x, B=2x\implies A+B=3x 1 You have a set of n numbers A = \{ x_1, \ldots, x_n \}, where we assume that those numbers are sorted: x_1 < x_2 < \cdots < x_n. You can form intervals [x_i, x_j] for i < j, with length x_{ij} = x_j - x_i. If one arranges the x_{ij} as a n\times n matrix, these are the elements above the diagonal. There are N = \frac{(n-1) ... 0 NO, there isn't. Consider : 1) \forall x \lnot \exists y P(x,y) 2a) \exists y \forall x P(y,x) 2b) \forall x \exists y P(y,x). Case I : consider 2a) and let a such that : \forall x P(a,x). By \forall-E we have : P(a,a). But from 1) we get the equivalent : \forall x \forall y \lnot P(x,y), and thus : \lnot P(a,a). ... 0 The "first statement" I interpret as \forall x\exists y \neg P(x,y) (as if we interpret it as you have written \forall x\neg\exists y P(x,y) then P needs to be the empty relation). As your formulation is vague and weird I have two choices of answers for you: *If I assume that a) is \exists y\forall x P(x,y) and b) is \forall x\exists y ... 1 For part (a), you need to prove subset inclusion in both directions; that is, f(A\diagdown C)\subseteq B\diagdown f(C) and f(A\diagdown C)\supseteq B\diagdown f(C). Here's how to write it up in full detail using proper mathematical language. (\subseteq) Let y\in f(A\diagdown C). Then y=f(a) for some a\in A\diagdown C. We know y\in B because ... 2 If f is continuous, then$$ f(\limsup_{n \to \infty}x_{n}) = f(\lim_{N \to \infty}\sup_{n \geq N}x_{n}) = \lim_{N \to \infty}f(\sup_{n \geq N}x_{n}); $$if in addition f is increasing, then f(\sup_{n \geq N}x_{n}) = \sup_{n \geq N}f(x_{n}) for all N \geq 1; hence$$\lim_{N \to \infty}f(\sup_{n \geq N}x_{n}) = \lim_{N \to \infty}\sup_{n \geq ...

1

A late answer; still: $\chi_E$ is continuous $\implies\forall x\in\Bbb R,\chi_E=1\ or\ \forall x\in\Bbb R,\chi_E=0$ Case 1: $\forall x\in\Bbb R,\chi_E=1\implies E=\Bbb R$ Case 2: $\forall x\in\Bbb R,\chi_E=0\implies E=\phi$ Both of which are open and closed (in fact, the only open and closed sets in $\Bbb R$)

0

This statement is extremely not true. It says ever single possible function must map, somehow by magic, to every single real number. So to show this is false all we have to do is show that some function doesn't map to some number. This is trivially easy. Any function that positive value function never has negative out put. Example $f(x,y) = (x + y)^2$. ...

1

$x+x-2x \equiv 0$ but $x^2+1=0$ is for only $x= \pm 1$. The first represents an identity which holds for all $x$ while the other is conditional equal which may or may not have solutions.

0

The sign you are referring to is the standard symbol for "identically equal to". For example, a more usual context in which to see that symbol (at least in pre-calculus classes) might be: $$(x-1)^{2} \equiv x^{2}-2x+1,$$ where the presence of the symbol indicates that the equation holds for all values of $x$. The reason for its use in this specific context ...

-3

Hint: If you choose sin x as your function what would you choose for $K$? $[-\pi/2, \pi/2]$?

4

That is false in general; consider the map $f: (x,y) \mapsto \sin x: \Bbb{R}^{2} \to \Bbb{R}$.

4

This is definitely not true. The easiest example is a constant function. $f(x,y)=c$ for all $(x,y)\in\Bbb R^2$. For a less trivial example, consider $f(x,y)=x^2+y^2$. Here $f(\Bbb R^2)=[0,\infty)$.

1

By the mean value theorem, any such function is a weak contraction. To find an example which is not a (strong) contraction, just pick any continuous function $g : \Bbb{R} \to (-1, 1)$ which satisfies $\lim_{x\to\infty} g(x) = 1$. Then $$f(x) = \int_{0}^{x} g(t) \, dt$$ will serve as a counter-example. For example, if we choose $g(x) = x^2/(1+x^2)$ then ...

0

Let $c>0$, $x\sin(1/x)$ is uniformly continuous on $[c/8,1]$ since $[c/8,1]$ is compact, this implies there exists $d>0$ such that for every $x,y\in [c/8,1]$, $\mid x-y\mid<d$ implies $\mid x\sin(1/x)-y\sin(1/y)\mid <c/4$, let $x,y\in (0,1)$ suppose that $\mid x-y\mid\leq inf(d,c/8)$ if $x,y>d/8, \mid x\sin(1/x)-y\sin(1/y)\mid <c/4$. If ...

2

As commenters have noted, the more common notation for this is $f^{-1}(C)$, which is denoted the pre-image or inverse image. If $f$ maps elements of a set $X$ to elements of a set $Y$, $f^{-1}$ maps subsets of $Y$ to subsets of $X$. The notation $f^*$ is used in the more abstract setting of sheaf theory, where it denotes the inverse image functor. In ...

1

$$\tan 3x = 3, \qquad\cot 2x=1/2$$ $$\tan 2x = 2$$ $$x = \frac{\arctan (2)}{2}$$ $$\tan\bigg[3\frac{\arctan(2)}{2}\bigg] = 3$$ This is not true, so the system does not hold

1

Hint: a continuous function on a closed bounded interval is uniformly continuous. A corollary of this is that if a continuous function can be continuously extended to a closed bounded interval it's uniformly continuous.

0

The function $f$ is onto: Case 1 Let $y \in [3,5)$ be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = 1 + 2^{1-n}$ is such that $f(x) = y$. In particular, if $n$ is such that $3 + 2^{1-n} \in [3,5)$ then $n \geq 0$. This in turn implies that $1 \leq 1 + 2^{1-n} \leq 2$. Case 2 Let $y \in [3,5)$ not be of the form: $3 + 2^{1-n}$ ...

0

The function $D(x)$ must have 2 second order zeros , so you must be able to write $$D(x)=a(x-b)^2(x-c)^2 \\ =ax^4-2a(b+c)x^3+ a(b^2+c^2 +4bc)x^2 +-2a(bc^2+b^2c)x + ab^2c^2$$ for real numbers $a,b,c$ from the definition of $D$ and using $L(x)=mx+d$ $$D(x)= -x^2(x+1)(x-2) - (mx+d) \\= -x^4 +x^3 +2x^2 - mx-d$$ Equating coefficients of $x$ gives you 5 ...

2

HINT: $D(x)$ is a quartic with two double roots, so it must have the form $D(x)=(x-r)^2(x-s)^2$, where $r$ and $s$ are the roots. On the other hand, if $L(x)=a+bx$, then $$(x-r)^2(x-s)^2=a+bx-2x^2-x^3+x^4\;,$$ so $$x^4-2(r+s)x^3+(r^2+4rs+s^2)x^2-2rs(r+s)x+r^2s^2=x^4-x^3-2x^2+bx+a\;.$$ Equate coefficients: \begin{align*} -2(r+s)&=-1\\ ... 1 if 2/(1+x)=1<->2=1+x<->x=1.. 1 If 2/(1+x) = 1, then by multiplying both sides by 1+x, you get 1+x=2. Subtracting 1 from both sides, you get x=1. 0 After some further investigation, it appears we can't actually get away with what's stated in the comments. Note that for a function f of two variables x and y, if y is also a function of x, we have that \frac{df}{dx}(x,y(x))=\frac{\partial f}{\partial x}(x,y(x))+\frac{\partial f}{\partial y}(x,y(x))\frac{dy}{dx}(x). $$It is worth pointing out ... 0 If X \stackrel{f}\to Y \stackrel{g} \to Z with g\circ f 1-1 then g may not be 1-1: other answers have given counterexamples. However, if f is surjective then g must be 1-1: if y_1,y_2\in Y with y_1\ne y_2, then y_i = f(x_i) for some x_i\in X, i = 1,2. We must have x_1\ne x_2, so g(f(x_1))\ne g(f(x_2)) — that is, g(y_1)\ne g(y_2. 2 This is not true, suppose f,g:\mathbb{R}\rightarrow \mathbb{R}, such that f(x)=e^x and g(x)=x^2. So g(f(x))=e^{2x} and therefore is 1-1, but g is not 1-1. 5 What happens if X=Z=\{1\}, Y=\{1,2\} and f:X\to Y is defined by f(1)=1 and g:Y\to Z is defined by g(1)=g(2)=1 ? 0 I put g(x)={\rm Arctan}(x). Note that g and its iterates g^{[n]}(x) are increasing. Hence for x\geq 1, we have g^{[n]}(x)\geq g^{[n]}(1)=u_n. The sequence u_n is positive, and satisfy u_{n+1}=g(u_n). As g(x)\leq x for all x, u_n is decreasing and u_n\to L, it is easy to show L=0. Now we can find a simple sequence w_n such that ... 1 Step 1. Applying the mean value theorem to the function x \mapsto x - \arctan x, for some \xi \in [0, A_{k-1}] we have$$ A_{k-1} - A_k = (A_{k-1} - \arctan A_{k-1}) = A_{k-1} \cdot \frac{\xi^2}{1 + \xi^2} \leq \frac{A_{k-1}^3}{1 + A_{k-1}^2}.$$This shows that$$ A_{k-1} - A_k \leq A_{k-1}^3 \qquad \text{and} \qquad \frac{A_{k-1}}{1+A_{k-1}^2} \leq ...

0

If $f:\mathbb{N}\rightarrow\mathbb{N}$ is increasing, then $g$ defined as $g(0)=f(0)$ and $g(n+1)=f(n+1)-f(n)$ can be any function $\mathbb{N}\rightarrow\mathbb{N}$. $f$ can be reconstructed from $g$ by $f(n)=\sum\limits_{k=0}^n g(k)$. Thus, this provides a bijection between increasing functions $\mathbb{N}\rightarrow\mathbb{N}$ and all functions ...

0

\begin{align}\\ & y'z'+x'y+x'yz+xyz'\\ &=y'z'+x'y(1+z)+xyz'\\ &=y'z'+x'y+xyz'\\ &=y'z'+(x'+xz')y\\ &=y'z'+(x'+z')y\\ &=y'z'+x'y+yz'\\ &=(y+y')z'+x'y\\ &=z'+x'y \end{align}

0

You are very close! To finish up, note that $$\bar z=\bar z(1+\bar x y),$$ from which you should see that answer 4 is correct.

0

The function $f(x)=x^3-3x^2-1$ is continuous on the whole real line, so you can apply the intermediate value theorem to any interval, for instance $[3.1,3.2]$; if you prove that $f(3.1)f(3.2)<0$, the theorem will tell you a zero exists in the open interval $(3.1,3.2)$. Rewrite $f(x)$ as $$f(x)=x^3-3x^2+3x-1-3x=(x-1)^3-3x$$ so $$... 0 Clearly f(x) is continuous in [3.1,3.2] and$$ f(3.1)f(3.2)<0$$by the Intermediate Mean Value Theorem, there is at least a c\in[3.1,3.2] such that f(c)=0. Note$$ f'(x)=3x(x-2)>0 in $[3.1,3.2]$, namely $f(x)$ is strictly increasing and therefore $c$ is a unique root in $[3.1,3.2]$.

0

Rule of signes: $f(x)=x^3-3x^2-1$ has $1$ change signe and $f(-1)=-x^3-3x^2-1$ has $0$ change signe, hence $f(x)=0$ has at least $3-(1+0)=2$ imaginary roots. Consequently the only real root is in your open interval because of $f(3)\lt 0$ and $f(3.2)\gt 0$.

2

If it's only real numbers you're working with, then no. If you're familiar with complex logarithms then $\log(z)=\log|z|+\mathrm{i}\theta$ might help, where $\theta$ is the argument of $z$.

2

By Descartes' Rule of signs, there is only one change in sign of coefficients of $f(x)$ and no change in sign of coefficients of $f(-x)$. So the equation $f(x)=0$ has exactly $1$ positive real root, no negative real root and $2$ complex conjugate roots. Now use $f(a)\cdot f(b)<0$ in the interval $[3.1,3.2]$ since the root does not lie either at $x=3.1$ ...

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