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10

The easy argument is that the two equations are incorrect because they violate the definition. The logarithm function permits a base that is strictly positive and not equal to one, and a domain that is strictly positive. Another way to see that your two examples can lead us to trouble is that you lose two very important properties of the logarithm. Namely, ...

9

You need every $x$ in the domain to have a $y$ in the codomain because it's a function, and that's the definition of a function. It had nothing to do with being injective or surjective.

8

Hint: take the graph of $x^3$, shift it one unit to the left, and then two units upwards.

8

They are correct in the sense that $(-3)^{2}=9$ and $(-2)^{3}=-8$. The difficulty arises because you want to think of $\log_{a}(x)$ as the inverse function of $a^{x}$. If $a>0$ (and $a\neq 1$) then $a^{x}$ is a continuous, real-valued (whenever $x$ is real), injective function (so $\log_{a}(x)$ makes sense). If $a=1$ then $a^{x}$ is not injective and ...

7

if you prove no real roots, in Fact we have $$(x+2a)^2+8a^2-8a+8=(x+2a)^2+8\left(a-\dfrac{1}{2}\right)^2+6>0$$ proof 2: $$\Delta =16a^2-4(12a^2-8a+8)=-32a^2+32a-32=-32\left(a-\dfrac{1}{2}\right)^2-24<0$$

6

No, consider $f,g:[0,\infty)\to[0,\infty)$ defined by $g(x)=x+1$ and $f(x)=\max\{0,x-1\}$. We have $fg=\operatorname{id}$ and $g(f(x))=\max\{1,x\}$, which is not the identity function. (For instance $g(f(0))=1.$) For finite sets, the answer is positive. In this case, $fg=\operatorname{id}$ implies that $g$ is injective and $f$ surjective. Since $X$ is ...

5

this is better done with logarithmic differentiation. here is how it works. we have $$y =\sqrt{\frac{2x-5}{3x+1}}\to \ln y = \frac12\left(\ln(2x-5) - \ln(3x+1)\right)$$ differencing the las equation we get $$\frac{2}{y}\frac{dy}{dx} = \frac{2}{2x-5} - \frac{3}{3x+1 } = \frac{17}{(2x-5)(3x+1)}$$ you can simplify more if it is needed.

5

Your initial step is correct, $f$ is continuous on $(0,1)$. I will assume the domain of $f$ is restricted to be $(0,1)$. Note that $\lim_{x \to 0^{+}} f(x)=\infty$ and $\lim_{x \to 1^{-}} f(x)=-\infty$, in other words near 0 and 1 the function takes arbitrarily large (or small) values. Since the function is continuous and defined on an interval, its image ...

4

With the given domain, $f$ is in fact one-to-one. But it is not onto the given codomain, because there is no $x$ with $f(x)=-1$. If the codomain is changed to $\mathbb{R}^+$, then $f$ is indeed invertible.

4

Its reduced discriminant is $$\Delta'=4a^2-(12a^2-8a+8)=-8(a^2-a+1)<0$$ as $\,a^2-a+1$ has no real roots (its roots are the complex cubic roots of $-1$). Alternatively, completing the square you can show $\,a^2-a+1\ge \dfrac34$.

4

Hint: What is $\frac{\mathrm{d}}{\mathrm{d}x}Ce^{-x}$, where $C$ is a real constant?

4

The tactic is to make things simpler. If there is $x$ such that $f(x)=[ln(\frac{7x-x^2}{12})]^\frac{3}{2}=1$ then we have also $ln(\frac{7x-x^2}{12})=1$, that is, $\dfrac{7x-x^2}{12}=e$ or even simpler $x^2-7x+12e=0$ You just have to check wether this equation has real solutions.

4

Think of a critical point as a "good candidate" for a point at which a local extremum could occur. To come up with a sensible way of formalizing this, think about common places where local extrema occur: for, say, $f(x)=x^2$, it's where $f'(x)=0$, but for $g(x)=|x|$, it's where $g'(x)$ is undefined (i.e. at $x=0$). We also want a critical point of a function ...

3

Where does the function $|x|$ attain it's minimum? Where does $\sqrt[3]{x}$ change concavity?

3

$$\frac{\sin^2 (\pi/2-x)}{\sqrt {\pi/2 -x}} = (\sin(\pi/2 -x))^{3/2}\cdot \left ( \frac{\sin (\pi/2-x)}{\pi/2 -x}\right )^{1/2} \to 0\cdot 1^{1/2} = 0.$$

3

It may help to let $\frac{\pi}{2}-x=t$. Then you will be on more familiar ground. For $t$ positive and not too large, we have $0\lt \sin t\lt t$. So our expression is positive and less than $t^{3/2}/\sqrt{2}$, which is less than $t$ if $t\lt 1$.

3

The formulation of the definition in the OP assumes existence and uniqueness of the inverse, and existence is not there always while uniqueness has to proved. I would reformulate it as: Let $\,f:X\to Y\,$ be a function. The function $\,g:Y\to X\,$ is said to be an inverse of $\,f$ if $\,g\circ f=i_X\,$ and $\,f\circ g=i_Y$. We denote the inverse of ...

3

This is the same as the equation $$1=\sum_{k=1}^{n}\frac{x^2}{x+a_k} = f(x).$$ Note $f$ is a nice $C^1$ function on $[0,\infty).$ Since $$f'(x) =\sum_{k=1}^{n}\frac{x^2+2xa_k}{(x+a_k)^2}>0, x> 0,$$ $f$ is strictly increasing on $[0,\infty).$ We have $f(0) = 0,$ and $f\to \infty$ at $\infty,$ so the intermediate value theorem shows $f$ takes on ...

3

Yes, there does. Let $$f(t)=\arctan(\pi t-\frac{\pi}{2})$$ This is the required function. Edit: As to how I came up with this function, it's something that everyone learns I suppose. A priori, you want a function to stretch out the interval to the whole real line, so "send" 1 to $\infty$ and $0$ to $-\infty$, as this is the most intuitive understanding for ...

3

You have right, from here you have to use this formula: $(u^n)'=n\cdot u^{n-1}\cdot u'$, where $u=(2x-5)(3x+1)^{-1}\Rightarrow u'=\frac{17}{(3x+1)^2}$ So you'll obtain: $\frac{1}{2} \cdot u^{-\frac{1}{2}} \cdot u'=\frac{1}{2}\cdot (2x-5)^{-\frac{1}{2}}(3x+1)^{\frac{1}{2}}\cdot\frac{17}{(3x+1)^2}=\frac{17}{2\sqrt{2x-5}(3x+1)^{\frac{3}{2}}}$

3

The issue I seewith your proof is that we haven't really established any form of contradiction. You can do it more directly by simply proving that composition of injective functions is injective. Which is done here

3

This is a fairly simple question, and questions like these are often asked by professors at the beginning of one's mathematical path, so that's what I'll assume. By that assumption, I will also presume that the level of strictness in your proof is similar to the typical level. For injectivity, you will need to be more specific. Did you prove a theorem ...

3

Suppose $\;g(z)\;$ is entire (i.e. analytic in the whole complex plane) and $\;g(z)=\dfrac1z\;$ on $\;|z|>1\;$ . Since $\;\dfrac1z\;$ is analytic on $\;0<|z|\le1\;$ as well, $\;g(z)$ continues analitically $\;\dfrac1z\;$ on the punctured unit disk. But since for any $\;z_k\to 0\;$ have that $\;g(z_k)\to g(0)\;$ and $\;g(0)\;$ is well defined, we get ...

2

Any function of the form $$y = ( a_1x-b_1)^2(a_2 x -b_2)^2 ( a_3 x - b_3)^2$$ will have your triple well form with centres $b_i/a_i$, this fixes 3 free variables. To determine the heights, you want the derivative at the peaks to be zero. Solve the resulting equations and you'll have your conditions. In your example we have $$y = x^2 (x-3)^2(x+3)^2$$ You ...

2

Hint: Take $x,y$ such that: $$(f\circ f \circ f)(x) = (f\circ f \circ f)(y).$$ Then $\color{blue}{f}$ injective gives: $$\color{blue}{f}(f(f(x))) = \color{blue}{f}(f(f(y)) \implies f(f(x)) = f(f(y)).$$ Repeat until you conclude that $x=y$.

2

For injectivity: Assume $f(x) = f(y)$. Either $x < y$, $x = y$ or $x > y$. If $x < y$ then $f(x) < f(y)$ by monotonicity, which is a contradiction. If $x > y$ then $f(x) > f(y)$ by monotonicity, which is a contradiction. So $x = y$. For surjectivity: Let $i \in [f(a),f(b)]$. By the intermediate value theorem, there is an $x \in [a,b]$ such ...

2

For the surjectivity it's not right. Let $y\in f([a,b])$. Then there is an $x\in[a,b]$ such that $y=f(x)$. By the fact that $g$ is increasing, $y\in[f(a),f(b)]\subset f([a,b])$. The fact that $[f(a),f(b)])$ is obvious by the intermediate value theorem.

2

I would suggest a small correction on the domain of your functions, namely to take $x\leqslant y <x^2- 2$ and to put "$p<x$" on the sum of $f_p(x)$. This way $t(x)$ seems correct, since it counts exactly for how many $y\in[x,x^2-2)$ holds $\text{lpf}(y(y+2)) > x$, and it clearly implies $y,y+2$ both primes, otherwise there would be a divisor of ...

2

You can also prove it directly. For simplicity let $Y^k$ be $Y$ nested $k$ times. We'll use the fact that function composition is associative. Suppose $Y^3(x)=Y^3(y)$ for $x,y\in X$. We can also write this as $Y(Y^2(x))=Y(Y^2(y))$. By injectivity of $Y$, it follows that $Y^2(x)=Y^2(y)$. In other words, $Y(Y(x))=Y(Y(y))$. Again by injectivity $Y$, this ...

2

Hint: This is one of the many functions from $S = \{ s_1, s_2, \ldots, s_6 \}$ to $\{ \alpha, \beta \}$: \begin{array}{c|c|c|c|c|c|c} s_i \in S & s_1 & s_2 & s_3 & s_4 & s_5 & s_6 \\ \hline f(s_i) & \beta & \beta & \alpha & \beta & \alpha & \alpha \end{array} This is a subset $A$ of $S$: \$A = \{ s_1, s_2, ...

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