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A subset E of X is called convex if, for any ##x,y \in E## and ##t \in (0,1)## then ##(1-t)x + ty \in E##. So by the inequality I wrote since ##\alpha f(x) + (1-\alpha)f(y)## is contained in the set it is convex?

Homework Statement [/B] Let X be a vector space over ##\mathbb{R}## and ## f: X \rightarrow \mathbb{R} ## be a convex function and ##g: X \rightarrow \mathbb{R}## be a concave function. Show: The set {##x \in X: f(x) \leq g(x)##} is convex. Homework Equations [/B] If f is convex...

Okay, I take from this because \sqrt{x_1}= \sqrt{x_2}, \sqrt{x_1}^2= \sqrt{x_2}^2, so x1=x2. So this function is one to one because I can prove that . Correct?

Homework Statement I am supposed to prove or disporve that ##f:\mathbb{R} \rightarrow \mathbb{R}## ##f(x)=\sqrt{x}## is onto. And prove or disprove that it is one to one Homework Equations The Attempt at a Solution I know for certain that this function is not onto given the codomain of real...

I updated it to define An.

Homework Statement For each ##n \in \mathbb{N}##, let ##A_{n}=\left\{n\right\}##. What are ##\bigcup_{n\in\mathbb{N}}A_{n}## and ##\bigcap_{n\in\mathbb{N}}A_{n}##. Homework Equations The Attempt at a Solution I know that this involves natural numbers some how, I am just confused on a...

Ah I understand, this makes more sense to me now. Thank you.

But one question: since i had ##\neg[(p\mid q)=x]## when the negation moves inside the expression it becomes: ##[(p\nmid q )\ne x]## Correct?

Oh thank you, correction made.

Homework Statement Express the following using existential and universal quantifiers restricted to the sets of Real numbers and natural numbers Homework Equations The Attempt at a Solution I believe the existence of rational numbers can be stated as: ##(\forall n \in \Re)(\exists p,q \in...

Homework Statement Express the following statement using only quantifiers. (You may only use the set of Real and Natural Numbers) 1. There is no largest irrational number. Homework Equations ##\forall=## for all ##\exists##=there exists The Attempt at a Solution I express the existence of...

Homework Statement Simplify the following statement as much as you can: (b). ##(3<4) \wedge (3<6)## Homework Equations ##\wedge= and## The Attempt at a Solution I figured that I could just write this as ##3<4<6##, but then I considered what if I didn't know that ##4<6## If it was just...

How about since ##\frac{x^2}{a^2}+\frac{y^2}{b^2}=1## ##(x-1)^2+y^2=1## Can I set each side equal to each other or should I solve for y^2 of the circle equation to plug into the ellipse equation.

I know this problem is emulative of https://www.physicsforums.com/threads/optimization-minimize-area-of-an-ellipse-enclosing-a-circle.270437/ this one however I am just getting into multivariable differentiation so this is very confusing to me.

I don't think I'm following that relationship either. How am I supposed to know the relationship between the ellipse and the circle.

Also how do we know that the circle and ellipse are in the range [0,2] for x values?

How exactly would I solve an auxiliary optimization to turn into a single restraint?

Homework Statement Consider the ellipse ##\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1## that encloses the circle ##x^{2}+y^{2}=2x##. Find the values of a and b that minimize the area of the ellipse. Homework Equations ##Area=ab\pi## The Attempt at a Solution I begin by completing the square...

Ok so ##\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}+C=0## ##C=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}## correct? So the beginning of this equation would be the magnitude of ##\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}+y_{0}-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}x_{0}-y_{0}## Over ##\sqrt{\left (...

I know it shouldn't be that hard, but I'm having such problems. So if the constant C is not zero how do I find it? From what I understand you saying I move all terms to one side to get: ##(y-y_{0})+\frac{\sqrt[3]{y}}{\sqrt[3]{x}}(x-x_{0})+c=0 ## So C is...

I still don't think I am understanding. Using the equation from wikipedia for distance: ##\frac{\left | ax_{0}+by_{0}+c \right |}{\sqrt{a^2+b^2}}## I can define "a" as the value: ##\frac{\sqrt[3]{y}}{\sqrt[3]{x}}## and "b" as the value: 1 Because in the slope form equation I have...

Ok the point slope equation would be: ##y-y_{0}=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\left ( x-x_{0} \right )##, Am I supposed to solve for Y0 and X0? How will this form be applied to finding the distance between the points?

If I put my equation in point slope form noting that the initial point is at (0,0) and I'm going to P(x,y) I would get y-0= ##\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}(x-0)## Which would be the same thing as y=##\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}x## But then how do you find distance when...

Ok using this equation, I went ahead and tried to find the distance to the specific point. Because the beginning is the origin. I just said that y0=Y-0 and x0=X-0 making the new equation for the line: ##y=\left ( -\frac{\sqrt[3]{y}}{\sqrt[3]{x}} \right )x##. And putting it into Ax+By+C form I...

So I did what you said and defined the tangent line as ##y=\left ( -\frac{\sqrt[3]{y_{o}}}{\sqrt[3]{x_{o}}} \right )x+b##. Does that look correct? And if so then how would I solve for "b" or even apply the distance formula to an equation defined this way?

Homework Statement Find the distance between the origin and the line tangent to ##x^\frac{2}{3}+y^{\frac{2}{3}}=a^{\frac{2}{3}}## at the point P(x,y) Homework Equations [/B] Distance= ##\frac{\left |a_{0}+b_{0}+c \right |}{\sqrt{a^{2}+b^{2}}}## The Attempt at a Solution To begin I find...

For differentiating the first equation I get##-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}##. And when I do the second derivative, and I end up with ##\left ( \frac{1}{3} \right )\sqrt[3]{\frac{y}{x^4}}##. Using these values I get ##\frac{3\left ( \frac{x^\frac{2}{3}+y^{\frac{3}{2}}}{x^{\frac{2}{3}}}...

When differentiating is it okay if I say that a^(2/3) will be a constant?

Does anyone have a hint? I have been working on this problem in between my breaks at school, and can't understand where to start.

Homework Statement Let T be the tangent line at the point P(x,y) to the graph of the curve ##x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}, a>0##. Show that the radius of curvature at P is three times the distance from the origin to the tangent line T. Homework Equations R=1/K...

I see what you mean. (I think) So I did what you said. ##3.56dx^{2}=(\frac{89 \sqrt{89}}{120})^{2}## ##dx=\frac{89}{24}## ##4-\frac{89}{24}=\frac{7}{24}## Using ##\frac{89}{24}##. I plug into dy to distance from y=1. ##dy=(\frac{89}{15})##. ##1+(\frac{89}{15})=\frac{104}{15}## The final...

Ok, thank you... I am not very confident with my reasoning, I feel like it could be right, but at the same time worried that I am making things up. Using the curvature equation I found that the radius of stated circle in the problem would be ##\frac{89\sqrt{89}}{120}##. Using this radius I...

Am I allowed to repost a question if I feel like I didn't get the help I needed?

I think this confusion comes from my lack of perentheses. My fault again. Or not what result are you getting? I check and still get the same thing ##\frac{\left (1+\left (\frac{5}{64}x^{\frac{3}{2}} \right )^{2} \right )^{3/2}}{\frac{15}{128}x^{\frac{1}{2}}}## Sorry if the notation is confusing.

So I kind of took what you said and reasoned with it. I noticed that center of the circle in the y direction would just be the radius so I just said (y-6.99)^2 for that part. Now for how much to shift the circle to the right or left I plugged in the value 4 for x and 1 for y so that I could see...

Sorry that was a very silly mistake when I differentiated incorrectly. ##y'=\frac{5}{64}x^{\frac{3}{2}}## ##y''=\frac{15}{128}x^{\frac{1}{2}}## So now ##\frac{((1+\frac{5}{64}x^{\frac{3}{2}})^2)^(3/2)}{\frac{15}{128}x^{\frac{1}{2}}}## Evaluating this at x=4 makes R=6.99 R^2=48.98 I...

Homework Statement A highway has an exit ramp that beings at the origin of a coordinate system and follows the curve ##y=\frac{1}{32}x^{\frac{5}{2}}## to the point (4,1). Then it take on a circular path whose curvature is that given bt the curve ##y=\frac{1}{32}x^{\frac{5}{2}}## at the point...

Thank you for the feedback. I will make sure to use latex ask questions about future problems!

Homework Statement Homework Equations Since this a proof. Most of the equations I needed were involved in the solution I created. The Attempt at a Solution I believe my proving is valid; however, I was wondering if I might have missed a step or did some math that didn't make sense,

Homework Statement An astronaut is pushed from her ship at a velocity of 2m/s. Her weight including her tool belt is 120kg. Remembering Newtons 3rd law, she takes 5 seconds to decide to detach her belt and throw it away from her. The 20kg tool belt leaves her suit as she throws it in front of...

Homework Statement When the thigh is fractured, the patient's leg must be kept under traction. One method of doing so is a variation on the Russell traction apparatus. If the physical therapist specifies that the traction force directed along the leg must be 25N, what must W be? Homework...

Homework Statement Today we were studying a block on an incline and determining the angle needed for the block to start sliding. The question asked is: Why is the equation for the coefficient of static friction independent of the weight of the block? Homework Equations ΣFx=Fapplied-fs=0...

Homework Statement This is a theoretical question from my homework. We did a lab in class where we were using motion to see how kinetic friction is affected by different variables. We found the acceleration of a sliding wooden block (sliding on another wooded block), and then used that...

What I have is ## \sin x+\sin (\frac{\pi}{2}-x)## using the Sum to Product Identity = ##\sqrt{ 2}\cos \frac {(\frac{\pi}{2}-x)}{2}## Putting that into our integral would be: ##\int \frac {\sin (\frac {\pi}{2}-x)}{\sqrt{ 2}\cos \frac {(\frac{\pi}{2}-x)}{2}}\ dx## =##\int \frac {\cos...

so for

I incidentally made that choice to leave those bounds. I now know that now though.

Homework Statement Use the substitution ##u=\frac{\pi} {2}-x## evaluate the integral ##\int_0^\frac {\pi}{2} \frac {\sin x}{\cos x + \sin x}dx##. Homework Equations [/B] ##\cos (\frac {\pi}{2}-x)=\sin x## The Attempt at a Solution [/B]I start by plugging "u" into the equation making the...