Tag Info

New answers tagged

5

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 ...


1

For each $x=(x_1,\ldots,x_n)\in\Bbb R^n$, we have that $f(x)$ is a vector in $\Bbb R^m$. Hence it has $m$-components, which we call $f_1(x),\ldots,f_m(x)$, i.e. $$f(x)=(f_1(x),\ldots,f_m(x))\in\Bbb R^m.$$ Each $f_i$ then gives a number in $\Bbb R$ for each $x\in\Bbb R^n$, that is $$f_i:\Bbb R^n\to\Bbb R.$$


1

Those $f_i$ are the $m$ component functions into which $f(x) \in \mathbb{R}^m$ can be decomposed. If $e_i$ is a canonical base vector of $\mathbb{R}^m$, we have $$ f(x) = \sum_i f_i(x)\, e_i \quad f_i(x) = f(x) \cdot e_i $$


1

We handle the modulus (absolute value sign), by splitting the function into two cases: $x<2$ and $x\ge 2$. The $x+2$ in the denominator means we also need to look at $x$ less-than and greater-than $-2$. So we look at three cases. Case 1: $x<-2$ Here $|x-2|=-(x-2)=2-x$ so $$f(x)=\frac{2-x}{x+2}=\frac 4{x+2}-1$$ As $x$ approaches $\infty$, $f(x)$ ...


1

Solving $f'(x) = 0$ we find that $f$ has a local maximum when $x = -1$, i.e. at $(-1,8-m)$ and a local minimum when $x = 1$, i.e. at $(1,-m)$. Also studying the sign of $f'$ we can conclude that $f$ increases for $x\in (-\infty,-1)\cup (1,+\infty)$ and decreases for $x\in (-1,1)$. Since $m\in (0,8)$ then the maximum is above $y=0$ and the minimum is below ...


0

Graphing the function shows that there are three distinct real roots to the equation $f(x) = m$ when $m \in (0,8)$. Now, to prove this formally, we utilise calculus. We have $f'(x) = 5x^4 - 5 = 5(x^4 - 1)$ and hence the turning points of the function are at $f'(x) = 0 \implies x = \pm 1$ which are the only real solutions to $f'(x) = 0$. Since $f(-1) = ...


0

Hint: note that $x^5$ is odd and make conclusions on the behavior of the function when $|x|$ is sufficiently large, show that $f(x)$ has two extremas, which satisfy that the maxima is $\geq 8$ and the minima $\leq 0$.


2

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

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 $n=4$


0

$\tan 0 = 0$, so use Maclaurin series expansion around 0 ( a couple of largest terms will do) and try plugging in values for $n$. What's the smallest one that gives you the result?


2

The graph is a parabola that opens upward. So there are two cases: $f(0) < 0$ f has a minimum at $x_0 < 0$ such that $f(x_0) < 0$ In the first case, $f(0) = a -3 < 0 \text{ gives } a<3$ In the second case, $f'(x) = 8x + a$ which has a minimum at $x_0 = -\frac{a}{8}$ where we require $a > 0$. Then $f(x_0) = \frac{a^2}{16} - ...


1

Examining the extrema of $f(x)=4x^2+ax+a-3$ which is a family of upwards open parabolas: $$ f'(x) = 8 x + a = 0 \Rightarrow x = -\frac{1}{8}a \\ f''(x) = 8 > 0 $$ So we have a local minimum at $x = -(1/8)a$. There the function has the value: $$ f(-(1/8)a) = \frac{1}{16}a^2 - \frac{1}{8}a^2 + a - 3 = -\frac{1}{16}a^2 + a - 3 $$ The minima regarding ...


3

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 !


3

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 ...


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 ...


0

The given expression is a quadratic in $x$ with the leading coefficient $\geq 0$. Since, it is given to be negative for at least one $x$, the conditions are: Discriminant $\geq 0$, that is $a^2-4a+12>0$ At least one root is negative. Solve for the roots and make the smaller one less than $0$.


0

You can treat this problem with roots indeed, for as soon as you have two distincts roots $x_1$ and $x_2$, all values between the two roots give a negative $f(x)$. This you can demonstrate easily. So to have distinct roots, you need to have a strictly positive discriminant, that is $a^2-4a+12>0$ You also want to have a negative root, that is $-a ...


2

The function will be positive when $x$ is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative. The smallest root is: $$\frac{-a -\sqrt{a^2-16(a-3)}}{8}$$ So we want to know when $a^2-16(a-3) > 0$, or $a^2-16a+48 > 0$, or $(a-8)^2-16 > 0$, or $(a-8)^2 > 16$, or ...


0

Let $f(x) = x+1$ and $g(x) = 5x+4$. Then, $$f \circ g = 5(x+1) + 4 = 5x + 9 = y$$ Thus, $$(f \circ g)^{-1} = 5y + 9 = x$$ Then, $$f^{-1}(x) = x-1$$ and $$g^{-1}(x) = \frac15(x-4).$$ So $$(f^{-1}\circ g^{-1}) = \frac15\left(x-1-4\right) = \frac15(x-5) = \frac15x - 1$$ Clearly $$\frac15x - 1 \neq 5y+9 = x.$$ Thus, $(f^{-1}\circ g^{-1}) \neq (f \circ ...


1

You should be able to get $[(f\circ g)^{-1} \circ (f \circ g)](x)=x $ and $[ (f\circ g) \circ (f \circ g)^{-1}](x)=x $ if you find the correct formula for $ (f \circ g )^{-1}$.


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!


1

Not quite. First you gave the correct formula $$ g'(x) = \frac{1}{f'(g(x))} \quad (*) $$ which by the way is a consequence of $f(g(x)) = x$ and taking the derivative of both sides. You then applied the wrong formula $$ g'(3) = \frac{1}{f'(3)} \mbox{vs.} \frac{1}{f'(g(3))} $$ which will not work in general except if $g(3) = 3$ which is not the case here: ...


1

To summarize what has been said about this: As Jonas Meyer noted in a comment, when you apply the inverse function the result has to be in the original function's domain. Since $-\frac{1}{\sqrt y} \not\in (0, \infty),$ that cannot be the correct answer, and there's only one possible answer remaining. In order to prove that that the only possible remaining ...


1

No. Suppose $f(10)=70$, so $g(70)=10$. And suppose $f'(10)=3$. It follows that $g'(70)=\dfrac 1 3$. It does not follow that $g'(10)=\dfrac 1 3$. You do just take reciprocals, but you need to evaluate each function at the right point.


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

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 $x=\frac{1}{\sqrt{y}}$.


0

Okay, well assuming I am understanding correctly your notation lets see what we can do. G is one-to-one , and thus invertible. That is, for the element -9, there exists only one element such that $g(x)=-9$ and that is $x=8$ For $h^{-1}$ we can do this systematically as follows, $h=4x-9$ , $h+9=4x$ $\frac{h+9}{4}=x$ and so this gives us that ...


1

It's not possible. Think of small $h> 0.$ Then by the MVT, $$\tag 1 [f(3R/4)-f(\,(R/2)+h\,)]/(\,(R/4)-h) = f'(c_h)$$ for some $c_h \in (R/2+h,3R/4).$ But $f(3R/4)=0,$ and because $f(R/2)=1,f'(R/2) = 0,$ we have $f((R/2)+h) = 1+o(h).$ So the left side of $(1)$ equals $$-(1+o(h)\,)/(\,(R/4)-h\,).$$ Now check that the absolute value of the above is ...


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.


0

The only Taylor expansions here are of $\cos(xy)$ and $\arctan(x^4)$. For $\cos(xy)$ in terms of $xy$ the error term in your approximation is simply $o(x^5y^5)$ as $\cos(xy)=\frac1{2!}(xy)^2-\frac1{4!}(xy)^4+\frac1{6!}(xy)^6+\dots$, and for $\arctan(x^4)$ our error term is $o(x^{11})$ as $\arctan(x^4)=x^4-\frac13(x^4)^3-\dots$


2

Given a particular function $f:X\to\{0,1\}$, let $S$ be the preimage of $1$. In other words, $$S=\{x\in X\mid f(x)=1\}$$ This is your desired bijection. For the inverse bijection, given a subset $S\subseteq X$, define $f:X\to\{0,1\}$ by $$f(x) = \begin{cases} 0, & \text{if $x\not\in S$} \\ 1, & \text{if $x\in S$} \end{cases}$$


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) = ...


1

how about $ f(x)=x^2\sqrt{100-x}$


0

If $y = 3x^2 - 2x^3 $, there are three main cases: $x$ near $0$; $x$ near $1$; other $x$. For $x$ near $0$, $x^3$ is small compared to $x^2$, so $y \approx 3x^2 $, so $x \approx \sqrt{y/3} $. For $x$ near $1$, let $x = 1-z$ where $z$ is small. Then $\begin{array}\\ y &=3(1-z)^2 - 2(1-z)^3\\ &=3(1-2z+z^2)-2(1-3z+3z^2-z^3)\\ ...


0

Notice that you cannot always do that,in this case you can;t. as it is not well defined. Assume that you can do that, and you can write $x=f(y)$. For $y=0$, correspond the values $x=0$ and $x=\frac{3}{2}$, so it's not well defined!


1

You want to start by supposing that $a,b\in A_i$ and $q_i(a)=q_i(b)$, i.e., that $\bar a=\bar b$, and your goal is to show that $a=b$. Since $\bar a=\bar b$, you know that there is some $k$ such that $i\le k$ and $p_{ik}(a)=p_{ik}(b)$. And therefore ... ?


1

$g(f(x)) = \frac{1}{q}$ x in rational $x = \frac{p}{q}$ in lowest terms otherwise $\sqrt{2}$. $f(g(x))= 0$ if $x = 0$ otherwise $\frac{1}{n}$ and $x=\frac{1}{n}$ in lowest terms


1

$f|_D: D\to A$ is a bijection where $A=\{{0}\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$ such that $f|_D^{-1}=g$ where $g: A \to D$ such that $$g(x)=\begin{cases} \frac1n,&x\in\Bbb Q\text{ and }x=\frac{1}n\text{ in lowest terms}\\ \sqrt{2},&x=0\ \end{cases}$$


2

That subset $D$ almost works, except you do not have anything that corresponds to $0$ which is in the range of $f$. Therefore you could add a particular irrational number to $D$. (A previous version of this answer said you could add zero, but that was a mistake.) So you can write the answer as $$D=\{y: y=\pi \lor y=\frac 1n \text{ for some }n\in\Bbb ...


2

You are supposed to map to the full range of $f$, which includes $0$. Hence you could take something like $D=\{\,\frac1n\mid n\in\mathbb N\,\}\cup\{\sqrt 2\}$


1

As the exponential function is strictly growing, the equation $2^x=y$ has a single real solution in $x$ (for $y>0$), which is called the logarithm (in base $2$), denoted as $\log_2(y)$. $$2^{\log_2(y)}=y.$$ The values of $2^x$ for increasing integers $x$ are $1,2,4,8,16,32\cdots$, and for negative integers, $\dfrac12,\dfrac14,\dfrac18\dfrac1{16}\cdots$, ...


0

The property you employed to conclude $x=3$ from $2^x = 2^3$ is called the injective property of the function $f(x)=2^x$. A function is said to be injective (or 1-1) if $$f(a)=f(b) \implies a=b.$$ In the case of $f(x) = n^x$ for a fixed positive $n$ (except for $n=1$), the function is injective, so you can make the same conclusion. Thus $3^x = 27 \implies ...


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 $y = a^x \iff x = \log_a y$.


0

Those cases are easy to solve if you have something like the example you just gave. Look for how does logarithm and exponential functions work. In other cases like $2^x=\pi$ that is more complicated to computate manually. By computate manually I mean by giving an explicit value in decimal notation. So look some info about exponential and logarithmic ...


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 ...


2

There are two things that could go wrong with your reasoning: If $f$ is defined for $x < -10$ or for $x > 13$, you have no idea how it behaves. You can't be sure that there is exactly one value of $x$ near $4$ so that $f(x)=15$. It could be the case that there is none, or one, or two. This depends on whether the value of the local maximum of $f(x)$ ...


1

$F(x)=f(|x+a|)-f(|x|)$ We have 3 cases. $1)$ $F(0) = 0 $ $2)$ $F(0) = f(a)-f(0) > 0 \implies F(-a) = f(0)-f(a) < 0$ $3)$ $F(0) = f(a)-f(0) < 0 \implies F(-a) = f(0)-f(a) > 0$ By composition of continuous functions F is continuous so we apply IVT and find $F(x_0)=0$ for some $ x_0 \in [-a,0] $ $F(x_0)=0 \implies f(|x_0+a|=f(|x_0|)$


1

I don't think that it is obvious that there is a real number $\alpha$ such that $3\lt \alpha\lt 5$ and $f(\alpha)=15$.


1

For sure, as pbs answered, adding the surface of the trapezoids is a way to go. However, you can do slightly better if, by chance, you have an odd number of data points. The method consists in fitting three data points by a parabola and integrate the resulting equation. Consider points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and you want the area between ...


1

Use the trapezoid rule. You can find information on this here: https://en.wikipedia.org/wiki/Trapezoidal_rule The integral can be interpreted as the area "under" the curve. You can approximate this area using the trapezoid rule, which will give you a value.



Top 50 recent answers are included