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8

You should visualize things on the real number line (Wikipedia link). The average of $x$ and $x'$, that is, the number $$\frac{x+x'}{2}$$ is exactly halfway between $x$ and $x'$. The distance between $x$ and $x'$ on the number line is $$|x-x'|$$ (regardless of which of $x$ and $x'$ is larger). Therefore, to get to the larger of the numbers $x$ and $x'$ ...


7

HINT: Start by finding an injection $f$ from $\Bbb R$ to the set of positive real numbers. Then the map $$g:\Bbb R\times\{0,1\}\to\Bbb R:\langle x,i\rangle\mapsto\begin{cases} f(x),&\text{if }i=0\\ -f(x),&\text{if }i=1 \end{cases}$$ is an injection. For $f$ consider an exponential function.


6

Let $f:A\rightarrow A$ with $f\circ f=1_A$ (The identity in $A$) If $f$ is not surjective then $f\circ f$ is not surjective and if $f$ is not injective $f\circ f$ is not injective. Hence if $f\circ f$ is bijective then $f$ is bijective. $f$ must therefore have an inverse $f^{-1}$. Using this we obtain the following: $f\circ f=1_a\implies f^{-1}\circ ...


5

What you suggest is sufficient for surjectivity. While it is true that you could not work out the decimal expansion of numbers like $\pi^2$ by hand in any finite amount of time, $\pi^2$ is a positive real number, so $\sqrt{\pi^2}=\pi$. The real numbers are unsettlingly complicated. Trying to work with the decimal expansions of real numbers can be messy, but ...


5

Having distinct real roots requires a positive discriminant, i.e. $$a^2-16(a-3)=(a-4)(a-12)>0,$$ or$$\color{green}{a<4\lor a>12}.$$ Then, having at least one negative root requires that either the sum or the product of the roots be negative, i.e. $$-a<0\lor a-3<0,$$which is always true and doesn't restrict the solution set !


5

Let $F(t) =\int f(t)dt , G(t) =\int g(t)dt $ then the your equation becomes $$\int f(t) g(t) dt = F(t) G(t) $$ and after differntiation $$F' G' =F'G+G'F $$ and hence $$1=\frac{G}{G' } +\frac{F}{F'} $$ if you take for example $$G(t) =\ln t $$ then you obtain $$1=\frac{\ln t}{\frac{1}{t} } +\frac{F}{F'} $$ hence $$\frac{F}{F'} =1-t\ln t$$ hence $$\ln F(t) = ...


4

Assume $f(y_1)=f(y_2)$. Then $y_1+f(1)=f(f(y_1)+1)=f(f(y_2)+1)=y_2+f(1)$ and hence $y_1=y_2$. Hence $f$ is injective. Given $a\in \mathbb R$, let $y=a-f(1)$. Then $f(f(y)+1)=a$. Hence $f$ is surjective.


4

You don't have to solve for the roots (which leads to irrational inequations!) to answer. Indeed, if $f(x)<0$ for at least one $x$, it has two real roots.The discriminant is $\;\Delta=a^2-16a+48=(a-4)(a-12)$, hence it has two real roots if $a<4$ or if $a>12$. Suppose $\Delta>0$; we'll determine the sign of the roots: $\;f(0)=a-3$, hence: if ...


4

Your last two formulas for $f(x)$ in your first edit are incorrect: they should be $$f(x)=1+\frac{6x}{x^2-3x+1}=1+\frac{6}{x-3+\frac{1}{x}}$$ (I see you corrected that formula in your question by a later edit.) You maximize that by making the denominator positive but as small as possible. The $-3$ in that denominator means that you make $x=3$. Any ...


4

$$x = \log_2 y \iff y = 2^x$$ The inverse is called the base two logarithm. In your case $2^x = 8 \iff x = \log_2 8 = \log_2 2^3 = 3$. In general the inverse for $a^x$, where $a> 1$ is the base $a$ logarithm. So $y = a^x \iff x = \log_a y$.


3

The composite function rule or the chain rule, same thing, state that if we have $$f(x)=h(g(x))$$ then $$f'(x) = h'(g(x)) \cdot g'(x)$$ Edit: To clarify, the "formula" for differentiating a composite function is the chain rule. Edit2: To further clarify, $h(g(x))$ is the same thing as $(h \circ g)(x)$. Just different notation.


3

Partial answer : If you want to find those, at least for integers, you can play around https://www.wolframalpha.com/input/?i=g%280%29%3Da%2C+g%28n%2B1%29%3Dn%5E4%2Bg%28n%29. You can find some functions provided you give $f(0)$, what I see from it is that it almost always exists!


3

Your polynomial is $p(x)=4x^2+ax+a-3$ its discriminant is $\Delta=a^2-16a+48$ and its roots are $\frac{-a\pm \sqrt{\Delta}}{8}$. Now there is a $x$ for which value is negative, and it is a parabola opening upwards, tells us that both roots are real and thus $\Delta>0 \implies a^2-16a+48=(a-4)(a-12)>0$ which happens iff $a\in ...


3

There is no real solution for the first question $1^x = 2$ Not even a complex number. For the second part: Let $x = a+ib$ and $e^{(2n+1)i\pi} = -1 , n \in \ Z $ so, $2^{a+ib} = 2e^{(2n+1)i\pi}$ $2^{a-1 + ib} = e^{(2n+1)i\pi}$ Take log and you will get, $(a-1+ib)\log2 = (2n+1)i\pi$ Equating real and imaginary part, $a = 1, b = \frac{(2n+1)\pi}{\log2}$ ...


3

Using Taylor series for $\tan(x)$ you get: $\tan(x)=x+\frac{x^3}{3} +O(x^5)$ $\tan^n(x)=(x+\frac{x^3}{3} +O(x^5))^n=x^n+\frac{n}{3}x^{n+2}+O(x^{n+4})$ This means that you can write your limit as: $$\lim_{x\to0} \frac{n}{3}\frac{x^{n+2}}{x^6}$$ If you want this limit to be a number you need to have $n+2 = 6$, thus $n=4$


3

Consider $g(x)=f(x)-m$. $g'(x)=5x^{4}-5=5(x^{4}-1)=5(x^{2}+1)(x^{2}-1)$. So g has extrema at $\pm 1$. Notice by the sign of $g'$, $g$ is increasing on $(-\infty,-1]$ and $[1,\infty)$ and decreasing on $[-1,1]$. $g(-1)=8-m$ and $g(1)=-m$. Now if $m \in (0,8)$, $g(-1)>0$ and $g(1)<0$ and you can use the fact that $g(x) \to \infty as x \to \infty$. Added ...


3

Take $y=\frac{x+2}{x+1}$ Then rearranging terms to get $x$ in the form of $y$. $$yx+y=x+2$$ $$yx-x=2-y$$ $$x(y-1)=2-y$$ $$x=\frac{2-y}{y-1}$$ As $x \geq 0$ $$\frac{2-y}{y-1} \geq 0$$ $$\frac{y-2}{y-1} \leq 0$$ $$y \in (1,2]$$


3

If $x>x'$, then $\frac{1}{2}(x+x'+|x-x'|)=\frac{1}{2}(x+x'+(x-x')=\frac{1}{2}(2x)=x=max(x,x')$. If $x<x', |x-x'|=x'-x$ and the same argument works and if $x=x'$ the result is obvious.


3

That the function is not defined on $a$, and thus it's not defined on $A$.


3

"Quadrilaterally" is probably what a confused person wrote when they meant "quadratically". And "monotonically" in the usage of mathematicians would simply mean that one of the quanitities, muscular torque, always increases as the other one, diameter, increases, as opposed to sometimes increasing and sometimes decreasing depending on the diamater. But in ...


3

If $\left \lfloor {x} \right \rfloor$ is the integer part, then for $0 \le x <1$ you have $\left \lfloor {x} \right \rfloor = 0$, so the limit towards $0^+$ is $0$ by the definition of $f$. For $-1 \le x < 0$ you have $\left \lfloor {x} \right \rfloor = -1$ so the limit towards $0^-$ is the limit of $\frac {\sin (-1)} {-1} = \sin 1$. Therefore, $f$ has ...


2

When you want to find an inverse function, the inverse must exist, ie the function must be bijective. Now since it is bijective on $(0, \infty)$, the inverse exists. Notice that $x>0$, so $x=\frac{1}{\sqrt{y}}$.


2

Your solution is not correct. For any rational number $a$ that is not itself an integer, observe that $f(a)=a$, but that for any $\epsilon<\lceil a\rceil-a$, there is no $\delta>0$ with the property that $$|x-a|<\delta\implies|f(x)-f(a)|=|f(x)-a|<\epsilon$$ because there exist irrational numbers $x$ arbitrarily close to $a$ with $f(x)=\lceil ...


2

That's not a redundant question at all! If I've understood you right, you want to know whether there could be an injection $A \rightarrow B$ and an injection $B \rightarrow A$ without there being a bijection between $A$ and $B$. Actually, this is not possible. There is a theorem called the Schröder–Bernstein theorem that states: Whenever there is an ...


2

Since you want to focus on finite sets, I will assume that a bag can only contain finitely many copies of each element. Let $\mathbb{N}_{> 0}$ be set of positive integers. Then a "bag-valued function" from $X$ to $Y$ is an ordinary map $X \to \mathbb{N}_{> 0} \times Y$. In order to define a category, we must give a composition law, and again, it ...


2

We need to find $g(3)$. $f(g(3))=3$, so $g(3)$ is a root of $a^{3}+3a-1=3$ so $a^{3}+3a-4=0$ so $(a-1)(a^{2}+a+4)=0$.By the quadratic formula, $a^{2}+a+4$ has no real root (discriminant is $-3<0$) so $g(3)=1$. Now $f'(x)=3x^{2}+3$ so $f'(g(3))=f'(1)=6$. So $g'(3)=\frac{1}{6}$. I hope I am not mistaken! Edited : I am assuming $g$ and $f$ are inverse ...


2

For determining the maximum of two arguments $x$, $x'$ we need the case distinction which of the two arguments is larger. For determining the absolute value of one argument $x$ one again needs a case distinction, here if the argument is negative or not. So it is not that surprising that one can express the maximum in terms of the absolute value. $$ ...


2

Actually $1\color{red}{\lt} f(x)\le 2$ for $x\ge 0$. To see this, note that we have $$\frac{x+2}{x+1}=\frac{(x+1)+1}{x+1}=1+\frac{1}{x+1}.$$ Here, note that this is strictly decreasing and that $\lim_{x\to\infty}\frac{1}{x+1}=0$.


2

In this case, it may be best to simply try your hand at a proof and then inspect it to see where you might be going wrong, if anywhere. For example: Claim: The mapping $f\colon\mathbb{R^+}\to\mathbb{R^+}$ defined by $f(x)=\sqrt{x}$ is bijective. Proof. We must show the mapping to be both injective and surjective: Injective: Suppose $f(x_1)=f(x_2)$. Then: ...


2

Below is a solution to this functional equation.



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