# Tag Info

0

The graph of the function $f(x) = x^2 - 3x + 4$ is the parabola $y = x^2 - 3x + 4$, which opens upwards since the leading coefficient is positive. Therefore, the function decreases to the left of the vertex, reaches its minimum value at the vertex, and increases to the right of the vertex. We can determine the vertex by completing the square. ...

0

When $b\in f(f^{-1}H)$ then exists $p\in f^{-1}H$ such that $f(p)=b$, but $f(p)\in H$ that is $b\in H$. Then $f(f^{-1}H)\subset H$. Now if $c\in H$ then by surjectivity exists $q\in A$ such that $f(q)=c$ that is $f(q)\in H$, so $q\in f^{-1}(c)\subseteq f^{-1}H$ then $f(q)\in f(f^{-1}H)$ that is $c\in f(f^{-1}H)$. Then $H\subset f(f^{-1}H)$.

1

Hint: let $h \in H$. Then since $f$ is surjective, there is some $a \in A$ such that $f(a) = h$. Then $a \in f^{-1}(H)$, so...

1

You can think of a vertical asymptote being a value of $x$ for which the function has a problem. Think of $\tan (90^{\circ})$ or $\frac 10$ - old calculators would say "Math error." Your function has a problem when the input value $x$ is equal to $5$. Your transformed function will have a problem when $2x+1=5$, which gives $x=2$. So the asymptote is now at ...

0

The problem just boils down to figuring out where $2x + 1 = 5$. You're just setting \begin{align*}\text{input of } f &= \text{location of old asympote}\\ 2x + 1 &= 5 \end{align*} since you know $f$ has an asymptote at $x = 5$. Remember that you can solve an equation in any order you'd like; the order of operations only tells you how to do ...

0

Let $y=2x+1$. Then $f(y)$ has an asymptote at $y=5$ (given). So the asymptote of $f(y)$ is at $$2x+1=y=5.$$ Solving, we find $x=2$.

0

Note that it isn't the same $x$. Therefore, if for $x_0$ there is an asymptote for $f(x),f(x_0)$, then this same $f(x)$ at point $x_1=2 x_0 + 2$, there is also the asymptote. So $x_0 = 5$ and $x_1=2x_0+1$, then $x_1 = 2$.

0

Write $$f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} = f_e(x) + f_o(x)$$ This means \begin{align} (f + g)(x) &= f(x) + g(x) \\ &= (f_e(x) + f_o(x) )+ (g_e(x) + g_o(x) ) \\ &= (f_e + g_e)(x) + (f_o + g_o)(x) \\ &= (f+g)_e(x) + (f+g)_o(x) \end{align} If $f = F_e$ and $g = G_o$ then \begin{align} (f+g)_e &= f_e + g_e = F_e + 0 = ...

1

You don't need smooth, or even continuous. Any function $f(x)$ is the sum of the two functions $$\frac{f(x) - f(-x)}{2}\quad,\quad \frac{f(x) + f(-x)}{2}$$ where the first one is odd and the second one is even. If $f$ itself is odd or even, then one of the fractions above evaluates to $0$ and the other to $f(x)$. These parts can be freely added to the ...

1

$Cont(g,f)=g\circ (\rm Id_X\times f)$. So it is not just similar to function composition, it is function composition.

0

In stead of $\{0,1,2,3\}^n$ let the domain of $f$ be a finite set $A$ with cardinality $m$. Exactly one element $a\in A$ must be selected to be sent to $1$. There are $m$ choices for $a$. For every element $b\in A-\{a\}$ there are $2$ choices for $f(b)$, resulting in $2^{m-1}$ possibilities in total. Final answer:$$m\times2^{m-1}$$ If $A=\{0,1,2,3\}^n$ ...

0

You choose which point $P$ of the domain is mapped into $1$ (you have $4^n$ choices), then you choose a function from $\{0,1,2,3\}^n\setminus\{P\}$ to $\{2,3\}$ (you have $2^{4^n-1}$ choices).

8

Note that $A := \{0,1,2,3\}^n$ has $4^n$ elements and $B := \{1,2,3\}$ has $2$ elements besides the 1. To give a function that takes $1$ exactly ones we first choose an element $a \in A$ which is mapped to $1$, there are $4^n$ possiblities for that, and after that, we have to choose images for the remaining $4^n - 1$ elements of $A \setminus \{a\}$, for each ...

1

Hint. Consider the function $f : [-1,1] \to \mathbb R$ defined by: $$\begin{cases}f(x) = x \sin \frac{1}{x} \text{ if } x \neq 0\\ f(0)=0\end{cases}$$ Considering $f$ on $[-1,0]$, what can be the value of $g(0)$? $f$ is continuous, but it is not of bounded variation, which forces a potential $g$ to take infinite values.

10

Let $f:[a, b]\to\mathbb R$ be a continuous function which is not of bounded variation, see here for an example. Then such a $g$ cannot be found. To see this, assume that $$|f(x) - f(y)| \le | g(x) - g(y)|$$ then by assumption, for any $M$, there is a partition $a =y_0< y_1 < y_2 < \cdots y_{n-1} < y_n = b$ so that $$\sum_{k=1}^n |f(y_{k}) - ... 0 One of the functions you have is 20, which is a constant, i.e. O(1). I think this is a good place to start, because all other functions are either o(1) or \omega(1). Also functions o(1) are fewer in numbers, so get those done first. 1 Here we have to calculate number of real values of x in x^2=x\sin x+\cos x So we will draw graph of f(x) = x^2 and f(x) = x\sin x+\cos x First one is upward parabola passing through origin and for drawing graph of f(x)=x\sin x+\cos x\;, we will put \displaystyle x=0\;,\frac{\pi}{2}\;,\pi,\frac{3\pi}{2},2\pi and f(-x) = f(x) Means f(x) is ... 2 Replace x by x \cos \theta - y \sin \theta and y by y \cos \theta + x \sin \theta where \theta is the angle to be rotated (in this case \theta = 25^\circ) 7 This is a nice, differentiable function, and the question asks for only the number of solutions. I'll use the standard method to attack problems of this kind: we try to divide the function up into areas where it is growing and falling, and then appeal to the intermediate value theorem over a few domains to get the answer. For this particular case, this ... 0 Your intuition seems to be wrong. For example, if f(x,y)=e^{-x^2-y^2} and g is any odd function of moderately growth at infinity, then all conditions are satisfied. 0 If f(x,y) = y g can be any integrable function that integrates to 0. If we take f(x,y) = y for x<0 and = y/2 for x>0 for example then we can take g to be anything that integrates to 0 using a measure of weight 1 for x > 0 and weight 2 for x < 0. So g can be pretty much anything if we make f really crazy. 2 The following does not use the fact that f is C^1. We assume only that f is even and f is differentiable at 0. Hint: Note that$$\frac{f(h) - f(0)}{h} = - \ \frac{f(-h) - f(0)}{(-h)}. $$Now recall that a limit exist if and only if both the left hand and right hand limit exists and are equal. 4 Your proof does not work, as h and x tend to 0 independently. Instead, differentiate both sides of the equation f(x) = f(-x), using the chain rule on the right hand side. You get f'(x) = -f'(-x). Then f'(0) = -f'(0), which implies f'(0) = 0. 0 The main difference that [ makes is, it complicates the example ;) First, an uncomplicated example: here's a bijection (0,1) \to (4,8):$$ x \mapsto 4 + 4x$$This first maps x \in (0,1) to 4x \in (0,4) – that itself is a bijection – and then (via another bijection) translates (0,4) to (4,8). You can biject any two open, closed, open-closed, or ... 0 Let f_0 be a bounded continuous nowhere-differentiable function. Define inductively f_n(x) = \int_0^x f_{n-1}(t) dt. So f_n \in C^n(\Bbb R) but is not n+1 times differentiable at any point. Define$$f(x) = \begin{cases} 0 & x = 0\\ e^{-1/x^2}f_n(x) & {1\over n+1} \le |x| < {1\over n},\; n \in \Bbb N_+\end{cases}$$This example is not ... 0 Hint: If f is not onto (surjective), then this relation on T need not be reflexive. For example: Let S=\{1,2,3\} with C=\{\{1\}, \{2,3\}\} and let T=\{a,b,c,d\}. Define f as$$f=\{(1,a), (2,b),(3,b)\}.$$Then c \not\sim c because f^{-1}(\{c\}) = \phi. By the definition of partition, \phi \not\subset C. Even if f is injective, say ... 0 The number of primes below a given number, \pi(x) can be estimated by: \pi(x) \approx \frac{x}{\ln(x)}. If you can find a "good" way to invert that function (and add maybe 20% uncertainty in there for small numbers), you can find the estimated upper bound. 1 You can prove that h(x) = ax^2 + bx + c is always positive by showing that the discriminant D=b^2-4ac is negative (thus showing that h has no real roots) and a>0 (thus showing that h is positive for large x). The two points together prove that, since h is continuous, h must be positive everywhere on \mathbb R. That said, you are way ... 1$$h(n)={2(4n^2+4n-3+|6n-3|)\over2n-1}$$will work for all non-negative integers. If this looks a little weird, here's where it came from: -1 I think I have found an answer:$$\frac{1}{\left \lceil\frac{1}{x}\right \rceil} \cdot (12 + 4x)$$Edit: I found a simpler answer without Infinity. 2$$h(n)=\left\{\begin{array}{ll}0&n=0\\12+4n&n>0\end{array}\right.\\ =(1-\text{sinc } \pi n)(12+4n)$$0 Um, the function is [\log_2 n ]  where [x] the least integer greater than to x. Since 2^n in binary is 1000....0 (with n zeros) it takes n + 1 = (\log_2 2^n) +1 bits to represent it. 2^{n+1} in binary is 100....0 with n + 1 zeros and n+2 bits. If m is such 2^n < m < 2^{n+1} it will require n+1 bits to represent it. As 2^n < m < ... 1 What you need to do is to solve for f(x) is the following: Convert the argument (the number, x) from decimal to binary. Count the number of digits. Return the number of digits in decimal form. How to convert to binary? An excellent primer is from Math Is Fun. Below is a link to the graphic on their website. You can convert a number to ... 0 By brute force: The values taken by q are$$1,6,11,4,9,2,7,0,5,10,3,8$$(repeatedly adding 5 and deducing 12 if necessary). Then by scanning the list in a loop by 5 elements, you find that 5\cdot5\equiv1\mod12, and$$y\equiv 5x+1\iff x\equiv5(y-1)\equiv5y+7.$$1 A necessary condition for a function to have an inverse is that the function is one-to-one. As there are only 12 elements in Z, you can find q(x) for all x \in Z and demonstrate that the function is one-to-one. 1 That is some poor wording on this question. (x^2-1)/(x-1) is not a function as domain/codomain are missing. It should better be written as$$f:\mathbb R\setminus\{1\}\rightarrow\mathbb R,~x\mapsto f(x)=\frac{x^2-1}{x-1}.$$For a function to be continuous at some point c of its domain, it is neccesary for the function to be defined at this point. As f ... 0 This type of relation occurs if one studies the linear response of a system to some cause. The standard example is the response of a dielectric to an electric field. There q is the electric field and p the susceptibility. But note that \begin{equation*} \int_{0}^{t}dsq(s)p(t-s)=\int_{0}^{t}dsq(t-s)p(s) \end{equation*} In general we have \begin{equation*} ... 1 The derivative of f on the interval (0,1) is$$ f'(x) = \frac{(x-1/2)^2 + 1/4}{x^2(1-x)^2}. $$Clearly, f'(x) > 0 for all x\in (0,1), so the function f is strictly increasing. This implies that the function f is injective on the interval (0,1), and even that it is bijective from (0,1) to its image. As f is continuous on the interval ... 0 Outline of proof: You have to show: 0<\dfrac{2b-2+\sqrt{4b^2+4}}{4b} < 1, and this is clear, and also (x-y)(2xy+1-x-y) = 0. If x = y then you're done, if not 2xy+1-x-y=0 \Rightarrow y = \dfrac{x-1}{2x-1}. You have to show that \dfrac{x-1}{2x-1} \notin (0,1) whenever x \in (0,1). 1 They tell you that it vanishes at exactly one point. But if f(x) is odd, then f(0) = 0. So f(\frac{1}{2}) > 0 by IVT. 4 The function is undefined at x = 1 because 0/0 is undefined. If you graphed this the graph would look like the graph of g(x) = x^2 with a point simply missing from the graph. We say that the limit of the function as x tends to x^2 (=1) from either x > 1 or x < 1 tends to 1. (\lim_{x \rightarrow 1} f(x) = x^2) We also refer to x=1 as a "removable ... 7 "without thinking what values variable takes..." - the answer to this is that you always should think about the values the variable(s) can take. A correct simplification is$$\frac{(x-1)x^2}{x-1}=x^2\ ,\quad\hbox{if $x\ne1$}.$$If you have a textbook which does not make the point that x\ne1, my suggestion is - throw it out and buy a better textbook! 0 Assuming that you are allowing real numbers for length and width, there are an infinite amount of possible solutions given the following: Area of a rectangle = length * width = 39 m^2  Perimeter of a rectangle =  2*length + 2*width  A(l,w) = lw = 39 P(l,w) = 2(l+w) 0 We know that Area = x * y where x and y are the side lengths of our rectangle. So, we can define a function to describe the perimeter: f(x,y) = 2(x+y) We know that Area = 39. I'll let you complete the rest. Hint: Solve for x in terms of y! 1 Rationals and irrationals are both dense in the real numbers. If we pick x\in\mathbb{Q} then there is a sequence of irrationals converging to x, thus proving discontinuity of f at x. If x\notin\mathbb{Q} then there is a sequence of rationals...(I'm sure you can now formally make your complete proof) 0 The second function does not contain z=2 in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers. 0 Here is an easier way to understand why if the limit of f(x) as x \to a is positive, then f(x) is positive on some neighborhood (a - \delta, a + \delta) around a. Let's call the limit of f(x) as x approaches a the letter L so that it's easy to talk about/reference. Suppose L > 0. Suppose on the otherhand that for every \delta > ... 0 Yes, one way of reading the definition of the derivative is as the existence of a function continuous at x=\xi which is \frac{f(x)-f(\xi)}{x-\xi} when x\neq \xi. There can only be one value of that function to make it continuous, and if that value is positive, you can use that theorem. It's useful to think with an actual constant in place of \xi. In ... 1 When x is irrational, the sequence defined in case (i) might not consist of only rationals. For example, if x = \sqrt 2, then x + \frac {\sqrt 2} n = \frac {n+1} n {\sqrt 2}  is irrational. (It will converge to x, but it doesn't accomplish what's needed.) For (ii), you need a sequence of rationals converging to the irrational x. In theory, we ... 0 I will suppose f(x) should be f(y) and f(z) and that we are working in \mathbb{R}. This is a rational expression. The only restriction is that the denominator is not 0. Since$$2y^2+1=0$$has no solution then the denominator does not impose any restriction on the domain. Hence dom(f)=\mathbb{R}. Same thinking leads to$$z-2\ne 0 \Leftrightarrow ...

Top 50 recent answers are included