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anonymous
 4 years ago
Prove that x^2 is always positive.
I understand why, but I don't know how to write out.
anonymous
 4 years ago
Prove that x^2 is always positive. I understand why, but I don't know how to write out.

This Question is Closed

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Try an indirect proof....assume that it is negative and show that it leads to a contradiction.

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0for every x in R y=x^2 stays equal to zero or bigger than zero

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5what do you mean you don't know how to write x^2>0?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5and yes because one counterexample for all complex would be if x=i i^2=1<0

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0correct....just look out because it could be equal to zero when x=0

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5so the best thing to say is x^2>=0 for all reals

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1show thats its minumm value is 0

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5which nenad already said

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1I butchered the crap out of that word....minimum ;)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5zarkon you don't have to know how to spell

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5you just have to get it a little right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here is the problem: f'(x) = (xa)(xb)^2 0 < a < b "For which values for a and b is f increasing?" And I can't motivate why (xb)^2 is always positive. Am I over thinking this?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5f'=2x=0 => x=0 which means x=0 is a critical number f''=2 which means it is concave up

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5so a and b are critical numbers \[f''=(xb)^2+(xa) \cdot 2(xb)=(xb)[xb+2(xa))]=(xb)[3xb2a]\] \[f''(a)=(ab)[3ab2a]=(ab)[ab]=(ab)^2=(ba)^2>0\] \[f''(b)=0 \text{ inconclusive }\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5b>a so b is not a ever so (ba)^2 is always positive of course we are pretending we have real a and real b

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5but it seems like the question you asking now is different from the one you are asking earlier \[f'(x)=(xa)(xb)^2\] we want f'>0 \[(xa)(xb)^2>0\] f' is zero when x=a or x=b  0 a b chose number btw 0 and a like a/2 we want the following to be greater than 0 \[f'(\frac{a}{2})=(\frac{a}{2})(\frac{a}{2}b)^2 \] we want to choose a number btw a and b now like (a+b)/2 again we want the following to be greater than 0 \[f'(\frac{a+b}{2})=\frac{a+b}{2}(\frac{ab}{2})^2=\frac{ab}{2}(\frac{ab}{2})^2=(\frac{ab}{2})^3\] now we choose a number after b like b+a we also want this to be bigger than 0 \[f'(b+a)=(b)(a)^2=a^2b\] no now we need to think of a and b such that all of these inequalities are satisfy

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5let me know if i made a mistake somewhere

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5so we have the following three inequalities: \[\frac{a}{2}(\frac{a}{2}b)^2>0, (\frac{ab}{2})^3>0 , a^2b>0\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5"For which values for a and b is f increasing?" wait i'm confused

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry I'll rephrase it

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5=> f is not increasing at x=a,x=b

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1that is not necessarly true

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5f'=0 when x=a f'=0 when x=b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry it's in Swedish but the picture might help.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5zarkon i want a counterexample from you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thats correct, x=a and x=b gives the zeros for the derivative

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5but we are to find values a and b so that f is increasing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes I know. And you get extra credit for pointing that out in the solution.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1def: a function \(f\) is increasing on and interval \(I\) if for any \(x,y\in I\) with \(x<y\) we have that \(f(x)<f(y)\) \[x^3\] satisfies this definition on\(\mathbb{R}\)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5but a, b cannot equal zero

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1I was just pointing out that just because the derivative is zero does not mean that the function is not increasing.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5right i get that part you are correct i didn't think about x^3

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0myininaya  as Zarkon says  f'(x)=0 could be a point of inflection. a function can be increasing/decreasing through a point of inflection

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5and other functions like x^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answers says that (xb)^2 will always be positive. Because of that (xa) determines the sign of the derivative. x > a and x != a,b Thats is all thats needed

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5but i wonder if i can think of one example where b>a>0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But the thing I couldn't solve was: Give a good motivation why (xb)^2 is always positive

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1the function is increasing over b the soluituion should just be x>a

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's not increasing if x=b or x=a

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1if you fix a value of x then f is not increasing(it is just a number)...you need to look over an interval....for some open set containig b the function is increasing...even though the derivative is zero at b

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0look at this as an example: http://www.wolframalpha.com/input/?i=y%3D%28x1%29%28x4%29%5E2+for+x%3D0+to+5 the function is increasing from x=4 onwards here 4 is equivalent to point x=b

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1i should say open interval

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1we ar given the derivative...graph the original function

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5http://www.wolframalpha.com/input/?i=find+a+and+b+such+that+a%5E2%2F2*%28a%2F2b%29%5E2%3E0%2C+%28ab%29%5E3%2F8%3E0%2C+a%5E2b%3E0 i put in those three inequalities wolfram said there is no solution to that system

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1this is never true for b>a>0 \[\frac{a}{2}(\frac{a}{2}b)^2>0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer doesn't require an interval. Even if x will equal a at some point, it's enough to state that at this point the derivative will be = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help Myininaya

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1you gave an interval (2 intervals) as an answer... " x > a and x != a,b Thats is all thats needed"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry sorry, I didn't think realize that. But that's the correct answer.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0myininaya  your original inequalities were correct  they lead to the correct solution of x > a

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5i don't see a solution to the problem

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0you got this for x < a:\[f'(\frac{a}{2})=(\frac{a}{2})(\frac{a}{2}b)^2\]which is always negative, so function cannot be increasing over this region. for x>a and x<b you got:\[f'(\frac{a+b}{2})=(\frac{ab}{2})^3=\frac{ab}{2}(\frac{ab}{2})^2=\frac{ba}{2}(\frac{ab}{2})^2\]which is always positive since b > a. and finally for x > b you got:\[f'(b+a)=a^2b\]which is always positive since 0 < a < b

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0therefore function is increasing for x > a

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5but want f'>0 always we wanted f to be always increasing and that gets messed up on my f'(a/2)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5oh we wanted to know where f is increasing?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5or did we want to choose a and b so that f is increasing everywhere?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.5first problem is x>a but second problem has no solution

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1it is not possible to pick an a and b such that f is increasing everywhere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is an analyzing problem. We should find when f is increasing and when it's zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We don't have to do that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just state that if x > a, f is increasing. as long as it isn't = a or b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But since it can't be > a and = a. x > a and x != b is the solution given in the answer sheet.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0I think your answer sheet is wrong, it should just be x > a as Zarkon stated earlier

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if x=b then f' = 0 which means not increasing

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0the function can still be increasing through a point of inflection

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1points do not increase and decrease...functions do over intervals

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0if you pick any specific point, then you cannot say that function is increasing/decreasing, it just has some value there. increasing/decreasing applies to an interval.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"The rate of change at a particular moment. Same as the value of the derivative at a particular point."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry I understand what you mean

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0talk about killing an ant with an h bomb

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not that i read the whole thing, but isn't is just asking for which values of x is \[xa>0\] namely \[x>a\]? if the function is continuous (and we assume it is since we assume it has a derivative everywhere) then the fact that the derivative may be postitive, then zero then positive means the funcion is increasing over the whole interval

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i see now that i read the posts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there was some confusion over what happens at b which reminds me
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